Matter Waves and de Broglie Wavelength Questions in English

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E$ is the kinetic energy.
Since the kinetic energy $E$ is the same for both the electron and the proton,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
We know that the mass of a proton $(m_p)$ is much greater than the mass of an electron $(m_e)$,i.e.,$m_p > m_e$.
Therefore,$\sqrt{m_p} > \sqrt{m_e}$,which implies $\frac{1}{\sqrt{m_p}} < \frac{1}{\sqrt{m_e}}$.
Thus,the de Broglie wavelength of the proton is less than that of the electron,i.e.,$\lambda_p < \lambda_e$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E$ is the kinetic energy.
Since the kinetic energy $E$ is the same for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{m_\alpha}{m_p}}$.
We know that the mass of an $\alpha$-particle $(m_\alpha)$ is approximately $4$ times the mass of a proton $(m_p)$,i.e.,$m_\alpha = 4m_p$.
Substituting this into the ratio: $\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m_p}{m_p}} = \sqrt{4} = 2$.
Thus,the ratio is $2:1$.

(B) The de-Broglie wavelength $\lambda$ for an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.
This implies $\lambda \propto \frac{1}{\sqrt{V}}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.
Given $\lambda_1 = 10^{-10} \ m = 1 \ \mathring{A}$,$V_1 = 150 \ V$,and $V_2 = 600 \ V$.
Substituting the values: $\frac{1 \ \mathring{A}}{\lambda_2} = \sqrt{\frac{600}{150}} = \sqrt{4} = 2$.
Thus,$\lambda_2 = \frac{1}{2} \ \mathring{A} = 0.5 \ \mathring{A}$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p = mv$ is the momentum of the particle.
Given: mass $m = 1 \ kg$,velocity $v = 1 \ m/s$.
Substituting these values into the formula:
$\lambda = \frac{h}{1 \times 1} = h$.
Therefore,the de Broglie wavelength of the ball is $h$.

(C) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.
Taking the derivative,we have $\frac{d\lambda}{dp} = -\frac{h}{p^2} = -\frac{\lambda}{p}$.
For small changes,we can write $\frac{\Delta \lambda}{\lambda} = -\frac{\Delta p}{p}$.
Taking the magnitude,we have $\left| \frac{\Delta \lambda}{\lambda} \right| = \left| \frac{\Delta p}{p} \right|$.
Given that the change in momentum $\Delta p = P_o$ and the percentage change in wavelength $\frac{\Delta \lambda}{\lambda} = 0.25\% = \frac{0.25}{100} = \frac{1}{400}$.
Substituting these values into the equation: $\frac{P_o}{p} = \frac{1}{400}$.
Therefore,the initial momentum $p = 400\ P_o$.

(B) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant $(6.6 \times 10^{-34} \ J \cdot s)$,$m$ is the mass of the electron $(9.1 \times 10^{-31} \ kg)$,and $v$ is the velocity.
Rearranging for velocity: $v = \frac{h}{m\lambda}$.
Substituting the given values: $v = \frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 5200 \times 10^{-10}}$.
Calculating the denominator: $9.1 \times 5200 \times 10^{-41} \approx 4.732 \times 10^{-37}$.
Thus,$v = \frac{6.6 \times 10^{-34}}{4.732 \times 10^{-37}} \approx 1.394 \times 10^3 \ m/s$.
Rounding to the nearest option,$v \approx 1.4 \times 10^3 \ m/s$.

(A) The de-Broglie wavelength $\lambda$ of a gas molecule at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.
This implies that $\lambda \propto \frac{1}{\sqrt{T}}$.
Given $T_1 = 27^oC = (27 + 273) K = 300 K$ and $T_2 = 927^oC = (927 + 273) K = 1200 K$.
Using the ratio: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{\lambda}{\lambda_2} = \sqrt{\frac{1200}{300}} = \sqrt{4} = 2$.
Therefore,$\lambda_2 = \frac{\lambda}{2}$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Given that the velocity $v$ becomes $2v$ and the kinetic energy $E = \frac{1}{2}mv^2$ becomes $2E$.
Let the new wavelength be $\lambda'$.
Using the relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,we have $\lambda \propto \frac{1}{\sqrt{E}}$.
If the energy $E$ is doubled to $2E$,the new wavelength $\lambda' = \frac{\lambda}{\sqrt{2}}$.
However,the question specifies that both velocity and energy are doubled. Since $E = \frac{1}{2}mv^2$,if $v$ is doubled,$E$ must become $4E$. If the problem states $E$ is doubled independently,it implies a change in mass or a specific constraint. Assuming the standard de Broglie relation $\lambda = \frac{h}{mv}$:
If $v' = 2v$ and $m$ is constant,then $\lambda' = \frac{h}{m(2v)} = \frac{\lambda}{2}$.
Given the options provided,the correct physical interpretation for a particle with constant mass is $\lambda' = \frac{\lambda}{2}$.

(C) According to the law of conservation of linear momentum,since the initial mass $M$ is stationary,the total initial momentum is $0$.
Therefore,the two fragments must move in opposite directions with equal magnitudes of momentum: $p_1 = p_2 = p$.
The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p}$.
Since both fragments have the same magnitude of momentum $p$,their de Broglie wavelengths will be $\lambda_1 = \frac{h}{p}$ and $\lambda_2 = \frac{h}{p}$.
Thus,the ratio of their de Broglie wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h/p}{h/p} = 1$.

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Since the wavelengths are equal,we have $\frac{h}{m_p v_p} = \frac{h}{m_e v_e}$.
This simplifies to $m_p v_p = m_e v_e$.
Given: mass of particle $m_p = 1 \, mg = 10^{-6} \, kg$,mass of electron $m_e = 9.1 \times 10^{-31} \, kg$,and velocity of electron $v_e = 3 \times 10^6 \, m/s$.
Substituting the values: $10^{-6} \, kg \times v_p = 9.1 \times 10^{-31} \, kg \times 3 \times 10^6 \, m/s$.
$v_p = \frac{27.3 \times 10^{-25}}{10^{-6}} \, m/s = 27.3 \times 10^{-19} \, m/s = 2.73 \times 10^{-18} \, m/s$.
Rounding to the nearest option,the velocity of the particle is $2.7 \times 10^{-18} \, m/s$.

(B) The de-Broglie wavelength $\lambda$ associated with an electron accelerated by a potential $V$ is given by the relation $\lambda = \frac{1.227}{\sqrt{V}} \; nm$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let $\lambda_1$ be the initial wavelength at $V_1 = 25 \; kV$ and $\lambda_2$ be the final wavelength at $V_2 = 100 \; kV$.
Taking the ratio,we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{100 \; kV}{25 \; kV}} = \sqrt{4} = 2$.
Therefore,$\lambda_2 = \frac{\lambda_1}{2}$.
This means the de-Broglie wavelength decreases by $2$ times.

(D) The radius $R$ of the circular path of a charged particle in a magnetic field is given by $R = \frac{mv}{Bq}$,which implies $mv = RBq$.
Given values:
$R = 0.83\, cm = 0.83 \times 10^{-2}\, m$
$B = 0.25\, Wb/m^2$
$q = 2e = 2 \times 1.6 \times 10^{-19}\, C$
Calculating the momentum $p = mv$:
$mv = (0.83 \times 10^{-2}) \times (0.25) \times (2 \times 1.6 \times 10^{-19})$
$mv = 0.83 \times 10^{-2} \times 0.25 \times 3.2 \times 10^{-19}$
$mv = 0.664 \times 10^{-21}\, kg\cdot m/s$
The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,where $h = 6.63 \times 10^{-34}\, J\cdot s$ (using $6.6$ for approximation as per the provided solution context):
$\lambda = \frac{6.6 \times 10^{-34}}{0.664 \times 10^{-21}} \approx 9.94 \times 10^{-13}\, m \approx 10^{-12}\, m$.
Converting to $\mathring{A}$ $(1\, \mathring{A} = 10^{-10}\, m)$:
$\lambda = 0.01 \times 10^{-10}\, m = 0.01\, \mathring{A}$.

(D) The de Broglie wavelength $\lambda$ is related to momentum $P$ by the equation $\lambda = \frac{h}{P}$.
Taking the logarithmic derivative,we get $\frac{d\lambda}{\lambda} = -\frac{dP}{P}$.
For small changes,we can write $\frac{|\Delta \lambda|}{\lambda} = \frac{\Delta P}{P}$.
Given that the wavelength changes by $0.5\%$,we have $\frac{\Delta \lambda}{\lambda} = 0.5\% = \frac{0.5}{100} = \frac{1}{200}$.
Substituting this into the relation,we get $\frac{\Delta P}{P} = \frac{1}{200}$.
Therefore,the initial momentum $P = 200\,\Delta P$.

(A) The de Broglie wavelength of an electron with kinetic energy $E$ is given by:
$\lambda_e = \frac{h}{\sqrt{2 m_e E}}$ .... $(i)$
The wavelength of a photon with energy $E$ is given by:
$\lambda_p = \frac{hc}{E}$ or $E = \frac{hc}{\lambda_p}$ .... $(ii)$
From equation $(i)$,squaring both sides:
$\lambda_e^2 = \frac{h^2}{2 m_e E}$ or $E = \frac{h^2}{2 m_e \lambda_e^2}$ .... $(iii)$
Equating the expressions for $E$ from $(ii)$ and $(iii)$:
$\frac{hc}{\lambda_p} = \frac{h^2}{2 m_e \lambda_e^2}$
Rearranging for $\lambda_p$:
$\lambda_p = \left( \frac{2 m_e c}{h} \right) \lambda_e^2$
Since $\frac{2 m_e c}{h}$ is a constant,we have:
$\lambda_p \propto \lambda_e^2$

(D) The de Broglie wavelength is given by the formula:
$\lambda = \frac{h}{\sqrt{2mK}}$ ...... $(i)$
where $m$ is the mass and $K$ is the kinetic energy of the particle.
When the kinetic energy of the particle is increased to $16$ times its initial value,the new kinetic energy $K' = 16K$.
The new de Broglie wavelength $\lambda'$ becomes:
$\lambda' = \frac{h}{\sqrt{2m(16K)}} = \frac{h}{4\sqrt{2mK}} = \frac{\lambda}{4}$ (Using equation $(i)$).
The percentage change in the de Broglie wavelength is calculated as:
$\text{Percentage change} = \frac{\lambda - \lambda'}{\lambda} \times 100$
$= \left(1 - \frac{\lambda'}{\lambda}\right) \times 100$
$= \left(1 - \frac{1}{4}\right) \times 100 = \frac{3}{4} \times 100 = 75\%$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electron is $K_{\max} = \frac{hc}{\lambda} - \phi_0$,where $\lambda$ is the wavelength of incident light and $\phi_0$ is the work function.
Given: $\lambda = 500\, nm$,$hc = 1240\, eV\, nm$,and $\phi_0 = 2.28\, eV$.
$K_{\max} = \frac{1240\, eV\, nm}{500\, nm} - 2.28\, eV = 2.48\, eV - 2.28\, eV = 0.2\, eV$.
The de Broglie wavelength $\lambda_e$ is related to kinetic energy $K$ by $\lambda_e = \frac{h}{\sqrt{2mK}}$. Since $K \le K_{\max}$,the minimum wavelength corresponds to the maximum kinetic energy: $\lambda_{\min} = \frac{h}{\sqrt{2mK_{\max}}}$.
Using $h = 6.6 \times 10^{-34}\, J\, s$,$m = 9.1 \times 10^{-31}\, kg$,and $K_{\max} = 0.2 \times 1.6 \times 10^{-19}\, J$:
$\lambda_{\min} = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 0.32 \times 10^{-19}}} \approx 2.8 \times 10^{-9}\, m$.
Since $K \le K_{\max}$,it follows that $\lambda_e \ge \lambda_{\min}$. Thus,$\lambda_e \ge 2.8 \times 10^{-9}\, m$.

(D) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ of a particle by the equation $\lambda = \frac{h}{p}$, where $h$ is Planck's constant.
This equation can be rewritten as $p \lambda = h$.
Since $h$ is a constant, the product of $p$ and $\lambda$ is constant $(p \lambda = \text{constant})$.
This relationship represents a rectangular hyperbola in the $p-\lambda$ plane, where $p$ decreases as $\lambda$ increases.
Therefore, the graph shown in Figure $D$ correctly represents this inverse relationship.

(D) For an electron of energy $E$,the de-Broglie wavelength is given by $\lambda_{e} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon of energy $E$,the relation is $E = \frac{hc}{\lambda_{p}}$,which implies $\lambda_{p} = \frac{hc}{E}$.
Taking the ratio of the wavelengths:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c} \times \frac{E^{1}}{\sqrt{E}} \times \frac{1}{\sqrt{2m}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{C}(\frac{E}{2m})^{1/2}$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Thus,$\lambda = \frac{h}{\sqrt{2mE}}$.
Given: $m = 1.7 \times 10^{-27} \; kg$,$E = 3 \; eV = 3 \times 1.6 \times 10^{-19} \; J$,and $h = 6.6 \times 10^{-34} \; J \cdot s$.
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1.7 \times 10^{-27} \times 3 \times 1.6 \times 10^{-19}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{16.32 \times 10^{-46}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{4.04 \times 10^{-23}}$
$\lambda \approx 1.633 \times 10^{-11} \; m$.
Rounding to the nearest option,we get $\lambda = 1.65 \times 10^{-11} \; m$.

(C) The kinetic energy $K$ of a neutron in thermal equilibrium at temperature $T$ is given by the equipartition theorem as:
$K = \frac{3}{2} k_B T$ ..... $(i)$
The relation between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$.
Substituting the value of $K$ from equation $(i)$:
$p = \sqrt{2m \cdot \frac{3}{2} k_B T} = \sqrt{3mk_B T}$
The de-Broglie wavelength $\lambda$ is given by:
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mk_B T}}$

(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.
The acceleration produced in the electron is $\vec{a} = \frac{\vec{F}}{m} = \frac{eE_0}{m} \hat{i}$.
The velocity of the electron at time $t$ is given by $\vec{v}_t = \vec{v} + \vec{a}t = \left(V_0 + \frac{eE_0}{m}t\right) \hat{i}$.
The magnitude of the velocity is $v_t = V_0 + \frac{eE_0}{m}t = V_0 \left(1 + \frac{eE_0}{mV_0}t\right)$.
The de-Broglie wavelength at time $t$ is $\lambda_t = \frac{h}{mv_t}$.
Substituting $v_t$,we get $\lambda_t = \frac{h}{m V_0 \left(1 + \frac{eE_0}{mV_0}t\right)}$.
Since the initial de-Broglie wavelength is $\lambda_0 = \frac{h}{mV_0}$,we have $\lambda_t = \frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$.

(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ accelerated through a potential difference $V$ is given by the formula:
$\lambda = \frac{h}{\sqrt{2meV}}$
where $h$ is Planck's constant and $e$ is the elementary charge.
Since $h$,$e$,and $V$ are constant for both particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
We know that the mass of an electron $(m_e)$ is much smaller than the mass of a proton $(m_p)$,i.e.,$m_e < m_p$.
Therefore,$\frac{1}{\sqrt{m_e}} > \frac{1}{\sqrt{m_p}}$,which implies $\lambda_e > \lambda_p$.
Thus,the correct option is $(C)$.

(D) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $v$ is its speed.
When an electron enters a uniform electric field,the electric force $F = qE$ acts on it,which can either accelerate or decelerate the electron depending on the direction of the field. Thus,the speed $v_1$ changes,causing $\lambda_1$ to change.
When an electron enters a uniform magnetic field,the magnetic force $F = q(v \times B)$ acts perpendicular to the velocity. This force changes the direction of motion but does not change the magnitude of the speed. Thus,the speed $v_2$ remains constant,and $\lambda_2$ remains constant.
Since the speed $v_1$ in the electric field can either increase or decrease relative to the initial speed,$\lambda_1$ can be either smaller or larger than $\lambda_2$.

(A) The initial kinetic energy of the electron is $K_i = 100 \, eV$.
When accelerated through a potential difference of $V = 50 \, V$,the electron gains additional kinetic energy equal to $K' = eV = 50 \, eV$.
The total kinetic energy of the electron becomes $K_f = K_i + K' = 100 \, eV + 50 \, eV = 150 \, eV$.
The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$.
Using the shortcut formula $\lambda (\text{in } \mathring{A}) = \sqrt{\frac{150}{K (\text{in } eV)}}$,we get:
$\lambda = \sqrt{\frac{150}{150}} = \sqrt{1} = 1 \, \mathring{A}$.

(B) Given: Mass of $A = m$,Mass of $B = \frac{m}{2}$. Initial velocity of $A = v$,Initial velocity of $B = 0$.
Let the final velocities be $v_1$ and $v_2$ for $A$ and $B$ respectively.
By conservation of momentum: $mv = mv_1 + (\frac{m}{2})v_2 \implies v = v_1 + \frac{v_2}{2} \implies 2v = 2v_1 + v_2$ ... $(i)$.
For an elastic collision,the coefficient of restitution $e = 1$,so $v_2 - v_1 = v - 0 \implies v_2 = v + v_1$ ... $(ii)$.
Substituting $(ii)$ into $(i)$: $2v = 2v_1 + (v + v_1) \implies v = 3v_1 \implies v_1 = \frac{v}{3}$.
Then $v_2 = v + \frac{v}{3} = \frac{4v}{3}$.
The de-Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Ratio $\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} = \frac{m_B v_2}{m_A v_1} = \frac{(\frac{m}{2}) \times (\frac{4v}{3})}{m \times (\frac{v}{3})} = \frac{\frac{2mv}{3}}{\frac{mv}{3}} = 2$.

(C) The initial de Broglie wavelength is $\lambda_0 = \frac{h}{m_0v_0}$.
When the wavelength becomes half,$\lambda = \frac{\lambda_0}{2} = \frac{h}{2m_0v_0}$.
Since $\lambda = \frac{h}{p}$,the final momentum $p$ must be $2m_0v_0$.
Let the velocity at time $t$ be $v$. Then $mv = 2m_0v_0$,so $v = 2v_0$.
The velocity components are $v_y = v_0$ (constant) and $v_x = a_x t = \frac{q_0E_0}{m_0}t$.
The magnitude of velocity is $v = \sqrt{v_x^2 + v_y^2}$.
Squaring both sides,$v^2 = v_x^2 + v_y^2 \Rightarrow (2v_0)^2 = v_x^2 + v_0^2$.
$4v_0^2 = v_x^2 + v_0^2 \Rightarrow v_x^2 = 3v_0^2 \Rightarrow v_x = \sqrt{3}v_0$.
Substituting $v_x = \frac{q_0E_0}{m_0}t$,we get $\frac{q_0E_0}{m_0}t = \sqrt{3}v_0$.
Therefore,$t = \sqrt{3} \frac{m_0v_0}{q_0E_0}$.

(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.
The wavelength of the first line of the Lyman series for a hydrogen-like ion with atomic number $Z$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For $He^+$,$Z = 2$. For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$.
$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R$.
Equating the two wavelengths: $\frac{h}{\sqrt{2meV}} = \frac{1}{3R}$.
Squaring both sides: $\frac{h^2}{2meV} = \frac{1}{9R^2}$.
Solving for $V$: $V = \frac{9R^2 h^2}{2me}$.

(A) The de Broglie wavelength $\lambda$ of an electron accelerated through a potential difference $V$ is given by the formula $\lambda = \frac{h}{\sqrt{2meV}}$.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{V}}$.
Let the initial wavelength be $\lambda = \frac{k}{\sqrt{V}}$ and the new wavelength be $\lambda^{\prime} = \frac{k}{\sqrt{4V}}$,where $k$ is a constant.
Dividing the two expressions,we get $\frac{\lambda^{\prime}}{\lambda} = \frac{\sqrt{V}}{\sqrt{4V}} = \frac{1}{2}$.
Therefore,$\lambda^{\prime} = \frac{\lambda}{2}$.
Comparing this with the given expression $\lambda^{\prime} = \frac{\lambda}{n}$,we find that $n = 2$.

(A) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Case $(i)$: The electron is moving opposite to the electric field. Since the electron is negatively charged, it experiences a force in the direction of the electric field, which accelerates it. As the velocity $v$ increases, the de-Broglie wavelength $\lambda$ decreases.
Case $(ii)$: The electron is moving perpendicular to a magnetic field. The magnetic force acts perpendicular to the velocity, changing the direction of motion but not the magnitude of the velocity $v$. Since the speed $v$ remains constant, the de-Broglie wavelength $\lambda$ remains the same.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}.$
Given that the wavelength changes by $0.50\%,$ the new wavelength $\lambda'$ is $\lambda' = \lambda - \frac{0.5}{100} \lambda = 0.995 \lambda = \frac{199}{200} \lambda.$
Since $\lambda = \frac{h}{p},$ the new momentum $p'$ is $p' = p + \Delta p.$
Thus,$\lambda' = \frac{h}{p + \Delta p}.$
Equating the two expressions for $\lambda':$
$\frac{h}{p + \Delta p} = \frac{199}{200} \left( \frac{h}{p} \right).$
$\Rightarrow \frac{1}{p + \Delta p} = \frac{199}{200p}.$
$\Rightarrow 200p = 199(p + \Delta p).$
$\Rightarrow 200p = 199p + 199 \Delta p.$
$\Rightarrow p = 199 \Delta p.$

(B) Let the masses of the two particles be $m$ and their speeds be $v$. The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$,so $mv = \frac{h}{\lambda}$.
In the centre of mass $(CM)$ frame,the velocity of particle $1$ is $\vec{v}_{1cm} = \vec{v}_1 - \vec{v}_{cm}$.
The velocity of the centre of mass is $\vec{v}_{cm} = \frac{m\vec{v}_1 + m\vec{v}_2}{2m} = \frac{\vec{v}_1 + \vec{v}_2}{2}$.
Thus,$\vec{v}_{1cm} = \vec{v}_1 - \frac{\vec{v}_1 + \vec{v}_2}{2} = \frac{\vec{v}_1 - \vec{v}_2}{2}$.
The magnitude of the relative velocity is $|\vec{v}_{1cm}| = \frac{1}{2} |\vec{v}_1 - \vec{v}_2|$.
Using the law of cosines for the vector subtraction of two vectors of equal magnitude $v$ at an angle $\theta = 60^o$:
$|\vec{v}_1 - \vec{v}_2| = \sqrt{v^2 + v^2 - 2v^2 \cos(60^o)} = \sqrt{2v^2 - 2v^2(0.5)} = \sqrt{v^2} = v$.
Therefore,$|\vec{v}_{1cm}| = \frac{v}{2}$.
The new de-Broglie wavelength is $\lambda' = \frac{h}{m|\vec{v}_{1cm}|} = \frac{h}{m(v/2)} = 2 \frac{h}{mv} = 2\lambda$.

(D) The initial momentum of the system is $P_i \approx 0$ because the neutron is moving slowly.
According to the law of conservation of linear momentum,the final momentum of the system must also be zero $(P_f = 0)$.
Let the momenta of the two product nuclei be $P_1$ and $P_2$. Since the total momentum is zero,we have $P_1 + P_2 = 0$,which implies $P_1 = -P_2$.
In terms of magnitude,$|P_1| = |P_2| = P$.
The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.
Since both nuclei have the same magnitude of momentum $P$,their de Broglie wavelengths will be $\lambda_1 = \frac{h}{P}$ and $\lambda_2 = \frac{h}{P}$.
Therefore,$\lambda_1 = \lambda_2 = \lambda$.

(C) Initial de Broglie wavelength of the electron is $\lambda_0 = \frac{h}{mv_0}$.
Force on the electron in the electric field is $\vec{F} = -e\vec{E} = -eE_0 \hat{j}$.
Acceleration of the electron is $\vec{a} = \frac{\vec{F}}{m} = -\frac{eE_0}{m} \hat{j}$.
The initial velocity is $\vec{v}_0 = v_0 \hat{i}$.
Velocity at time $t$ is $\vec{v} = \vec{v}_0 + \vec{a}t = v_0 \hat{i} - \frac{eE_0 t}{m} \hat{j}$.
The magnitude of velocity at time $t$ is $v = \sqrt{v_0^2 + (\frac{eE_0 t}{m})^2} = v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}$.
The de Broglie wavelength at time $t$ is $\lambda = \frac{h}{mv} = \frac{h}{m v_0 \sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.
Substituting $\lambda_0 = \frac{h}{mv_0}$,we get $\lambda = \frac{\lambda_0}{\sqrt{1 + \frac{e^2 E_0^2 t^2}{m^2 v_0^2}}}$.

(D) The initial kinetic energy of the particle is $K_i = E$.
The initial de-Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK_i}} = \frac{h}{\sqrt{2mE}}$.
When the particle enters a region with potential energy $V$,its total energy remains conserved. The new kinetic energy $K_f$ is given by $K_f = E - V$.
The new de-Broglie wavelength $\lambda'$ is $\lambda' = \frac{h}{\sqrt{2mK_f}} = \frac{h}{\sqrt{2m(E - V)}}$.
We can rewrite this as $\lambda' = \frac{h}{\sqrt{2mE(1 - \frac{V}{E})}} = \frac{h}{\sqrt{2mE}} \cdot \frac{1}{\sqrt{1 - \frac{V}{E}}}$.
Substituting $\lambda = \frac{h}{\sqrt{2mE}}$,we get $\lambda' = \lambda \left( 1 - \frac{V}{E} \right)^{-\frac{1}{2}}$.

(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ is related to momentum $p$ by $\lambda = \frac{h}{p}$.
For a particle in a box of length $L$ with perfectly elastic collisions,the standing wave condition requires that the length $L$ must be an integer multiple of half-wavelengths: $L = \frac{n\lambda}{2}$,where $n = 1, 2, 3, \dots$.
Rearranging for $\lambda$,we get $\lambda = \frac{2L}{n}$.
Substituting this into the de-Broglie relation: $p = \frac{h}{\lambda} = \frac{h}{(2L/n)} = \frac{nh}{2L}$.

(B) For diffraction to occur in the nickel granules,the condition is $2d \sin \theta = n \lambda$.
Since $\sin \theta \le 1$,the maximum wavelength that can be diffracted is $\lambda_{\max} = 2d$ (for $n=1$).
Electrons with wavelengths $\lambda > 2d$ cannot be diffracted by the nickel granules and will pass through the pipe.
Using the de Broglie relation $\lambda = \frac{h}{p}$,the condition $\lambda > 2d$ implies $\frac{h}{p} > 2d$,or $p < \frac{h}{2d}$.
The kinetic energy $E$ of these electrons is $E = \frac{p^2}{2m}$.
Substituting the inequality for $p$,we get $E < \frac{(\frac{h}{2d})^2}{2m} = \frac{h^2}{8md^2}$.
Thus,the electrons that pass through the pipe have energy less than $\frac{h^2}{8md^2}$.

(B) Given that the de-Broglie wavelength of the particle is equal to the wavelength of the photon: $\lambda_p = \lambda_{ph} = \lambda$.
For a photon,the energy is $E_{ph} = \frac{hc}{\lambda}$.
For a particle,the de-Broglie wavelength is $\lambda = \frac{h}{mv}$,which implies $h = \lambda mv$.
The kinetic energy of the particle is $K_p = \frac{1}{2}mv^2$.
Substituting $h$ into the photon energy equation: $E_{ph} = \frac{(\lambda mv)c}{\lambda} = mvc$.
The ratio of the kinetic energy of the particle to the energy of the photon is:
$\frac{K_p}{E_{ph}} = \frac{\frac{1}{2}mv^2}{mvc} = \frac{v}{2c}$.
Given $v = 2.25 \times 10^8\, m/s$ and $c = 3 \times 10^8\, m/s$:
Ratio $= \frac{2.25 \times 10^8}{2 \times 3 \times 10^8} = \frac{2.25}{6} = \frac{225}{600} = \frac{3}{8}$.

(B) The de-Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V$ is given by the relation $\lambda = \frac{h}{\sqrt{2mqV}}$.
This implies that $\lambda \propto \frac{1}{\sqrt{V}}$.
Given $\lambda_1 = 10^{-10} \ m = 1 \ \mathring{A}$ at $V_1 = 150 \ V$.
We need to find $\lambda_2$ at $V_2 = 600 \ V$.
Using the ratio: $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{V_2}{V_1}}$.
Substituting the values: $\frac{1}{\lambda_2} = \sqrt{\frac{600}{150}} = \sqrt{4} = 2$.
Therefore,$\lambda_2 = \frac{1}{2} = 0.5 \ \mathring{A}$.

(C) The de Broglie wavelength $\lambda$ of a charged particle accelerated through a potential difference $V_0$ is given by $\lambda = \frac{h}{\sqrt{2mqV_0}}$.
Since $h$ and $V_0$ are constant for both particles,the wavelength is inversely proportional to the square root of the product of mass and charge: $\lambda \propto \frac{1}{\sqrt{mq}}$.
For a proton,$m_p = m$ and $q_p = e$. For an $\alpha$-particle,$m_{\alpha} = 4m$ and $q_{\alpha} = 2e$.
The ratio of the wavelengths is $\frac{\lambda_p}{\lambda_{\alpha}} = \sqrt{\frac{m_{\alpha}q_{\alpha}}{m_pq_p}} = \sqrt{\frac{4m \times 2e}{m \times e}} = \sqrt{8} = 2\sqrt{2}$.
Thus,the ratio is $2\sqrt{2} : 1$.

(C) At the topmost point,the velocity of the bomb of mass $M$ is $v = u \cos \theta$ in the horizontal direction. The momentum is $P = M u \cos \theta$.
After the explosion,the bomb splits into two fragments of mass $m = M/2$. Let the velocity of the first fragment (which returns to the point of projection) be $v_1$ and the second fragment be $v_2$.
Since the first fragment returns to the point of projection,it must have a horizontal velocity of $-u \cos \theta$ (as it travels the same horizontal distance back in the same time).
By conservation of linear momentum in the horizontal direction:
$M u \cos \theta = (M/2)(-u \cos \theta) + (M/2) v_2$
$u \cos \theta = -0.5 u \cos \theta + 0.5 v_2$
$1.5 u \cos \theta = 0.5 v_2 \implies v_2 = 3 u \cos \theta$.
The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
For the bomb just before explosion: $\lambda_{bomb} = \frac{h}{M u \cos \theta}$.
For the second fragment just after explosion: $\lambda_2 = \frac{h}{(M/2) v_2} = \frac{h}{(M/2) (3 u \cos \theta)} = \frac{2h}{3 M u \cos \theta}$.
The ratio of the de Broglie wavelength of the second fragment to that of the bomb is:
$\frac{\lambda_2}{\lambda_{bomb}} = \frac{\frac{2h}{3 M u \cos \theta}}{\frac{h}{M u \cos \theta}} = \frac{2}{3}$.

(C) The de-Broglie wavelength $\lambda$ of an electron with kinetic energy $K$ is given by $\lambda = \frac{h}{\sqrt{2mK}}$.
Given $K = 1.5 \ eV = 1.5 \times 1.6 \times 10^{-19} \ J$.
Using the formula $\lambda = \frac{12.27}{\sqrt{V}} \ \mathring{A}$,where $V$ is the accelerating potential in volts. Since $K = eV$,$V = 1.5 \ V$.
$\lambda = \frac{12.27}{\sqrt{1.5}} \ \mathring{A} \approx \frac{12.27}{1.225} \ \mathring{A} \approx 10 \ \mathring{A}$.
Converting to meters: $\lambda = 10 \times 10^{-10} \ m = 10^{-9} \ m$.

(A) The de Broglie wavelength $\lambda$ of an electron with energy $V$ (in volts) is given by $\lambda = \sqrt{\frac{150}{V}} \, \mathring{A}$.
Given energy $E = 10 \, KeV = 10^4 \, eV$,so $V = 10^4 \, V$.
$\lambda = \frac{12.27}{\sqrt{10^4}} \, \mathring{A} = \frac{12.27}{100} \, \mathring{A} = 0.1227 \, \mathring{A}$.
According to Bragg's law for diffraction,$n \lambda = d \sin \phi$,where $d = 0.2454 \, \mathring{A}$ and assuming $n = 1$ for the first order.
$0.1227 = 0.2454 \sin \phi$.
$\sin \phi = \frac{0.1227}{0.2454} = 0.5$.
$\phi = \arcsin(0.5) = 30^{\circ}$.

(B) For a proton,the de Broglie wavelength is given by $\lambda_1 = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
For a photon,the energy is given by $E = pc$,so the momentum is $p = \frac{E}{c}$. The de Broglie wavelength is $\lambda_2 = \frac{h}{p} = \frac{hc}{E}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_1}{\lambda_2} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} \times \frac{E}{hc} = \frac{1}{c\sqrt{2m}} \times \frac{E}{E^{1/2}} = \frac{1}{c\sqrt{2m}} \times E^{1/2}$.
Thus,$\frac{\lambda_1}{\lambda_2} \propto E^{1/2}$.
Therefore,the correct option is $B$.

(C) The de-Broglie wavelength $\lambda$ is related to the kinetic energy $K$ of an electron by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant and $m$ is the mass of the electron.
Let the initial kinetic energy be $K_1 = K$ and the initial wavelength be $\lambda_1 = \frac{h}{\sqrt{2mK}}$.
If the kinetic energy doubles,the new kinetic energy is $K_2 = 2K$.
The new wavelength is $\lambda_2 = \frac{h}{\sqrt{2m(2K)}} = \frac{h}{\sqrt{2} \cdot \sqrt{2mK}} = \frac{1}{\sqrt{2}} \lambda_1$.
Therefore,the de-Broglie wavelength changes by a factor of $\frac{1}{\sqrt{2}}$.

(B) The total energy $E$ of the particle is given as $2E_0$.
For the region $0 \le x \le 1$,the potential energy is $U(x) = E_0$. The kinetic energy $K_1$ is $E - U = 2E_0 - E_0 = E_0$.
The de-Broglie wavelength is $\lambda_1 = \frac{h}{\sqrt{2mK_1}} = \frac{h}{\sqrt{2mE_0}}$.
For the region $x > 1$,the potential energy is $U(x) = 0$. The kinetic energy $K_2$ is $E - U = 2E_0 - 0 = 2E_0$.
The de-Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2mK_2}} = \frac{h}{\sqrt{2m(2E_0)}} = \frac{h}{\sqrt{4mE_0}}$.
Taking the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2mE_0}}}{\frac{h}{\sqrt{4mE_0}}} = \sqrt{\frac{4mE_0}{2mE_0}} = \sqrt{2}$.
Therefore,$(\lambda_1/\lambda_2)^2 = (\sqrt{2})^2 = 2$.