De Broglie's principle Questions in English

(A) According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2 \pi r = n \lambda$.
For the $n^{th}$ Bohr orbit, the radius is given by $r = n^{2} a_{0}$, where $a_{0}$ is the Bohr radius.
For the $4^{th}$ orbit $(n = 4)$, the radius is $r = 4^{2} a_{0} = 16 a_{0}$.
Substituting these values into the quantization equation: $2 \pi (16 a_{0}) = 4 \lambda$.
Solving for $\lambda$: $\lambda = \frac{32 \pi a_{0}}{4} = 8 \pi a_{0}$.

(A) According to the de Broglie equation: $\lambda = \frac{h}{mv}$
Given: $h = 6.626 \times 10^{-34} \,J \,s$,$m = 0.1 \,kg$,$v = 10 \,m \,s^{-1}$
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \,J \,s}{(0.1 \,kg)(10 \,m \,s^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{1} \,m = 6.626 \times 10^{-34} \,m$

The kinetic energy $(K.E.)$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.
First,calculate the velocity $(v)$:
$v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 3.0 \times 10^{-25} \ J}{9.1 \times 10^{-31} \ kg}} \approx 812 \ m \ s^{-1}$.
Now,use the de Broglie wavelength formula: $\lambda = \frac{h}{mv}$.
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{(9.1 \times 10^{-31} \ kg) \times (812 \ m \ s^{-1})}$.
$\lambda \approx 8.967 \times 10^{-7} \ m = 896.7 \ nm$.

Given wavelength $\lambda = 3.6 \, \mathring{A} = 3.6 \times 10^{-10} \, \text{m}$.
Velocity of photon $v$ is equal to the speed of light $c = 3 \times 10^8 \, \text{m s}^{-1}$.
Using the de Broglie relation $m = \frac{h}{\lambda v}$,where $h = 6.626 \times 10^{-34} \, \text{J s}$:
$m = \frac{6.626 \times 10^{-34} \, \text{J s}}{(3.6 \times 10^{-10} \, \text{m})(3 \times 10^8 \, \text{m s}^{-1})}$
$m = \frac{6.626 \times 10^{-34}}{10.8 \times 10^{-2}} \, \text{kg}$
$m = 6.135 \times 10^{-33} \, \text{kg}$.

(A) According to de Broglie's equation,$\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \,J s$
$m = 9.109 \times 10^{-31} \,kg$
$v = 2.05 \times 10^{7} \,m s^{-1}$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34}}{(9.109 \times 10^{-31})(2.05 \times 10^{7})}$
$\lambda = \frac{6.626 \times 10^{-34}}{18.673 \times 10^{-24}}$
$\lambda = 0.3548 \times 10^{-10} \,m = 3.548 \times 10^{-11} \,m$.

(N/A) From de Broglie's equation,$\lambda = \frac{h}{mv}$.
Given,Kinetic energy $(K.E.)$ of the electron $= 3.0 \times 10^{-25} \ J$.
Since $K.E. = \frac{1}{2} mv^{2}$,therefore velocity $(v) = \sqrt{\frac{2 K.E.}{m}}$.
$v = \sqrt{\frac{2(3.0 \times 10^{-25} \ J)}{9.1 \times 10^{-31} \ kg}} = \sqrt{6.5934 \times 10^{5}} \approx 812 \ ms^{-1}$.
Substituting the value in the expression of $\lambda$:
$\lambda = \frac{6.626 \times 10^{-34} \ Js}{(9.1 \times 10^{-31} \ kg)(812 \ ms^{-1})} \approx 8.96 \times 10^{-7} \ m$.
Hence,the wavelength of the electron is $8.96 \times 10^{-7} \ m$.

From de Broglie's equation,$\lambda = \frac{h}{mv}$.
Given,Kinetic energy $(K.E)$ of the electron $= 3.0 \times 10^{-25} \ J$.
Since $K.E = \frac{1}{2} mv^2$,therefore velocity $(v) = \sqrt{\frac{2 K.E}{m}}$.
$v = \sqrt{\frac{2(3.0 \times 10^{-25} \ J)}{9.1 \times 10^{-31} \ kg}} = \sqrt{6.5934 \times 10^5} \approx 812 \ ms^{-1}$.
Substituting the value in the expression of $\lambda$:
$\lambda = \frac{6.626 \times 10^{-34} \ Js}{(9.1 \times 10^{-31} \ kg)(812 \ ms^{-1})} \approx 8.96 \times 10^{-7} \ m$.
Hence,the wavelength of the electron is $8.96 \times 10^{-7} \ m$.

(N/A) According to Bohr's postulate,the angular momentum of an electron in a hydrogen atom is given by:
$mvr = n \frac{h}{2\pi}$ ........$(1)$
where $n = 1, 2, 3, \dots$
According to de Broglie's equation,the wavelength $\lambda$ is given by:
$\lambda = \frac{h}{mv}$
Rearranging this gives:
$mv = \frac{h}{\lambda}$ ........$(2)$
Substituting the value of $mv$ from equation $(2)$ into equation $(1)$:
$r \left( \frac{h}{\lambda} \right) = n \frac{h}{2\pi}$
Canceling $h$ from both sides and rearranging:
$2\pi r = n\lambda$ ........$(3)$
Since $2\pi r$ represents the circumference of the Bohr orbit,equation $(3)$ proves that the circumference of the Bohr orbit is an integral multiple of the de Broglie wavelength associated with the electron.

According to de Broglie's equation, $\lambda = \frac{h}{mv}$.
Given:
$h = 6.626 \times 10^{-34} \,Js$
$m = 9.109 \times 10^{-31} \,kg$
$v = 1.6 \times 10^{6} \,ms^{-1}$
Substituting the values:
$\lambda = \frac{6.626 \times 10^{-34} \,Js}{(9.109 \times 10^{-31} \,kg)(1.6 \times 10^{6} \,ms^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{1.4574 \times 10^{-24}} \,m$
$\lambda = 4.546 \times 10^{-10} \,m \approx 4.55 \times 10^{-10} \,m$
Converting to picometers:
$\lambda = 455 \,pm$.

According to de Broglie's equation, $\lambda = \frac{h}{mv}$.
Rearranging for velocity, we get $v = \frac{h}{m\lambda}$.
Given:
$h = 6.626 \times 10^{-34} \, J \cdot s$
$m = 1.67493 \times 10^{-27} \, kg$ (mass of a neutron)
$\lambda = 800 \, pm = 800 \times 10^{-12} \, m$
Substituting the values:
$v = \frac{6.626 \times 10^{-34}}{(1.67493 \times 10^{-27}) \times (800 \times 10^{-12})}$
$v = \frac{6.626 \times 10^{-34}}{1.339944 \times 10^{-36}}$
$v \approx 494.5 \, m \cdot s^{-1}$.
Thus, the characteristic velocity associated with the neutron is approximately $494.5 \, m \cdot s^{-1}$.

According to de Broglie's equation, $\lambda = \frac{h}{mv}$.
Where:
$\lambda = \text{wavelength associated with the electron}$
$h = 6.626 \times 10^{-34} \, J s \text{ (Planck's constant)}$
$m = 9.109 \times 10^{-31} \, kg \text{ (mass of electron)}$
$v = 2.19 \times 10^{6} \, m s^{-1} \text{ (velocity of electron)}$
Substituting the values in the expression of $\lambda$:
$\lambda = \frac{6.626 \times 10^{-34} \, J s}{(9.109 \times 10^{-31} \, kg)(2.19 \times 10^{6} \, m s^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{1.9948 \times 10^{-24}} \, m$
$\lambda \approx 3.32 \times 10^{-10} \, m$
Since $1 \, pm = 10^{-12} \, m$, we have:
$\lambda = 332 \times 10^{-12} \, m = 332 \, pm$.
Therefore, the de Broglie wavelength associated with the electron is $332 \, pm$.

According to de Broglie's expression,$\lambda = \frac{h}{mv}$.
Substituting the given values:
$h = 6.626 \times 10^{-34} \, J \cdot s$
$m = 0.1 \, kg$
$v = 4.37 \times 10^{5} \, ms^{-1}$
$\lambda = \frac{6.626 \times 10^{-34} \, J \cdot s}{(0.1 \, kg)(4.37 \times 10^{5} \, ms^{-1})}$
$\lambda = \frac{6.626 \times 10^{-34}}{4.37 \times 10^{4}} \, m$
$\lambda = 1.516 \times 10^{-38} \, m$

(N/A) de-Broglie principle: The French physicist,de-Broglie,in $1924$ proposed that matter,like radiation,should also exhibit dual behaviour,i.e.,both particle and wavelike properties.
This means that just as the photon has momentum as well as wavelength,electrons should also have momentum as well as wavelength.
Analogy of de-Broglie principle: The following relation exists between wavelength $(\lambda)$ and momentum $(p)$ of a material particle:
$\lambda = \frac{h}{mv} = \frac{h}{p} \quad (Eq. -2.30)$
Proof of de-Broglie's dual behaviour: de-Broglie's prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction,a phenomenon characteristic of waves.
Limitation: According to de-Broglie,every object in motion has a wave character. The wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.
The wavelengths associated with electrons and other subatomic particles (with very small mass) can,however,be detected experimentally.

(N/A) de-Broglie principle: The French physicist,de-Broglie,in $1924$ proposed that matter,like radiation,should also exhibit dual behaviour,i.e.,both particle and wavelike properties. This means that just as the photon has momentum as well as wavelength,electrons should also have momentum as well as wavelength.
Mathematical relation: The following relation exists between wavelength $(\lambda)$ and momentum $(p)$ of a material particle:
$\lambda = \frac{h}{m v} = \frac{h}{p} \quad$ (Eq. $- 2.30$)
Proof of de-Broglie's dual behaviour: de-Broglie's prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction,a phenomenon characteristic of waves.
Limitation: According to de Broglie,every object in motion has a wave character. The wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.
The wavelengths associated with electrons and other subatomic particles (with very small mass) can,however,be detected experimentally.

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
For an electron,the velocity $v$ is given by $v = \frac{h}{m \lambda}$.
Given: $h = 6.626 \times 10^{-34} \ J \cdot s$,$m = 9.11 \times 10^{-31} \ kg$,and $\lambda = 5200 \ \mathring{A} = 5.2 \times 10^{-7} \ m$.
Substituting the values:
$v = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31}) \times (5.2 \times 10^{-7})}$
$v = \frac{6.626 \times 10^{-34}}{47.372 \times 10^{-38}}$
$v = 0.1398 \times 10^4 \ m/s \approx 1400 \ m/s$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant $(6.626 \times 10^{-34} \ J \cdot s)$,$m$ is the mass of the proton $(1.67 \times 10^{-27} \ kg)$,and $K$ is the kinetic energy in Joules.
First,convert kinetic energy from $eV$ to Joules: $K = 500 \ eV \times 1.602 \times 10^{-19} \ J/eV = 8.01 \times 10^{-17} \ J$.
Now,substitute the values into the formula: $\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 8.01 \times 10^{-17}}}$.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2.675 \times 10^{-43}}} = \frac{6.626 \times 10^{-34}}{1.635 \times 10^{-22}} \approx 4.05 \times 10^{-12} \ m$.

The velocity of the electron is $v = (\frac{1}{100}) \times (3.00 \times 10^8 \ m \ s^{-1}) = 3.00 \times 10^6 \ m \ s^{-1}$.
The momentum of the electron is given by $p = m \times v$.
Substituting the mass of the electron $(m = 9.11 \times 10^{-31} \ kg)$:
$p = (9.11 \times 10^{-31} \ kg) \times (3.00 \times 10^6 \ m \ s^{-1}) = 2.733 \times 10^{-24} \ kg \ m \ s^{-1}$.
The de-Broglie wavelength is calculated using the formula $\lambda = \frac{h}{p}$,where $h = 6.626 \times 10^{-34} \ J \ s$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{2.733 \times 10^{-24} \ kg \ m \ s^{-1}} = 2.424 \times 10^{-10} \ m$.

Given: $KE = 4.55 \times 10^{-25} \ J$,$m = 9.1 \times 10^{-31} \ kg$.
$1$. Velocity $(v)$:
$KE = \frac{1}{2} mv^2$
$4.55 \times 10^{-25} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2$
$v^2 = \frac{2 \times 4.55 \times 10^{-25}}{9.1 \times 10^{-31}} = 10^6$
$v = 10^3 \ m \ s^{-1}$.
$2$. Momentum $(p)$:
$p = mv = 9.1 \times 10^{-31} \ kg \times 10^3 \ m \ s^{-1} = 9.1 \times 10^{-28} \ kg \ m \ s^{-1}$.
$3$. Wavelength $(\lambda)$:
Using de Broglie equation,$\lambda = \frac{h}{p}$
$\lambda = \frac{6.626 \times 10^{-34} \ J \ s}{9.1 \times 10^{-28} \ kg \ m \ s^{-1}} \approx 7.28 \times 10^{-7} \ m$.

(A) $1$. Kinetic Energy $(KE)$: $KE = e \times V = (1.602 \times 10^{-19} \ C) \times (1.0 \times 10^4 \ V) = 1.602 \times 10^{-15} \ J$.
$2$. Frequency $(
u)$: Using $KE = h\nu$,$\nu = \frac{KE}{h} = \frac{1.602 \times 10^{-15} \ J}{6.626 \times 10^{-34} \ J \cdot s} = 2.418 \times 10^{18} \ Hz$.
$3$. Wavelength $(\lambda)$: Using de Broglie equation $\lambda = \frac{h}{\sqrt{2m(KE)}}$,$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-15}}} = 1.227 \times 10^{-11} \ m$.

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.
Rearranging for momentum: $p = \frac{h}{\lambda}$.
Given: $\lambda = 1 \ \mathring{A} = 10^{-10} \ m$ and $h = 6.625 \times 10^{-34} \ J \ s$.
Substituting the values: $p = \frac{6.625 \times 10^{-34} \ J \ s}{10^{-10} \ m} = 6.625 \times 10^{-24} \ kg \ m/s$.

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Given that the distance traveled in $1 \ s$ is equal to the wavelength,we have $d = v \times t = v \times 1 = v$.
Thus,$\lambda = v$.
Substituting this into the de Broglie equation: $v = \frac{h}{mv}$,which implies $v^2 = \frac{h}{m}$.
Here,$h = 6.626 \times 10^{-34} \ J \ s$ and $m = 9.11 \times 10^{-31} \ kg$.
$v^2 = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31}} \approx 7.27 \times 10^{-4} \ m^2 \ s^{-2}$.
Taking the square root,$v = \sqrt{7.27 \times 10^{-4}} \approx 2.7 \times 10^{-2} \ m \ s^{-1}$.

(A) Given: Kinetic Energy $(KE) = 4.55 \times 10^{-25} \ J$,mass of electron $(m) = 9.1 \times 10^{-31} \ kg$,Planck's constant $(h) = 6.626 \times 10^{-34} \ J \ s$.
Using the formula for kinetic energy: $KE = \frac{1}{2} mv^2$.
$4.55 \times 10^{-25} = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2$.
$v^2 = \frac{2 \times 4.55 \times 10^{-25}}{9.1 \times 10^{-31}} = \frac{9.1 \times 10^{-25}}{9.1 \times 10^{-31}} = 10^6$.
$v = 10^3 \ m \ s^{-1}$.
Applying the de Broglie equation: $\lambda = \frac{h}{mv}$.
$\lambda = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^3} = \frac{6.626}{9.1} \times 10^{-6} \ m \approx 0.728 \times 10^{-6} \ m = 7.28 \times 10^{-7} \ m$.

Given: Mass $m = 100 \ g = 0.1 \ kg$.
Velocity $v = 100 \ km/h = \frac{100 \times 1000 \ m}{3600 \ s} = 27.78 \ m/s$.
Planck's constant $h = 6.626 \times 10^{-34} \ J \cdot s$.
According to the de Broglie equation,$\lambda = \frac{h}{mv}$.
Substituting the values: $\lambda = \frac{6.626 \times 10^{-34} \ J \cdot s}{0.1 \ kg \times 27.78 \ m/s} = 2.385 \times 10^{-34} \ m$.
The calculated wavelength is extremely small $(2.385 \times 10^{-34} \ m)$,which is far beyond the range of any detectable measurement. Therefore,the wave nature of macroscopic objects like a cricket ball is not observable.

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.
Since the wavelength $\lambda$ is the same for both particles,the product $mv$ must be constant $(mv = \frac{h}{\lambda} = \text{constant})$.
This implies that $v \propto \frac{1}{m}$.
Since the mass of an electron $(m_e \approx 9.11 \times 10^{-31} \ kg)$ is much smaller than the mass of a proton $(m_p \approx 1.67 \times 10^{-27} \ kg)$,the electron must have a higher velocity to maintain the same wavelength.

(A) The de Broglie principle states that all matter,especially microscopic particles like electrons,exhibits dual behavior,i.e.,both particle-like and wave-like properties.
The relationship is given by the equation: $\lambda = \frac{h}{mv}$,where $\lambda$ is the wavelength,$h$ is Planck's constant,$m$ is the mass of the particle,and $v$ is its velocity.

(B) The phenomenon of diffraction is a characteristic property of waves. Since an electron beam undergoes diffraction,it indicates that electrons possess wave-like properties,as proposed by the $De \ Broglie$ hypothesis.

(N/A) An electron microscope can achieve a magnification of up to $15$ million times.
Principle: It is based on the wave nature of electrons,as proposed by the de Broglie hypothesis,where moving electrons behave as waves with a very short wavelength.

(B) The wave nature of ordinary objects cannot be detected.
According to the de Broglie equation,$\lambda = \frac{h}{mv}$.
Since the mass $(m)$ of ordinary objects is very large,the wavelength $(\lambda)$ associated with them is extremely small.
This wavelength is too small to be measured or detected by any experimental means.

(A) Using the de Broglie equation: $\lambda = \frac{h}{mv}$.
Given: $h = 6.626 \times 10^{-34} \ J \cdot s$,$m = 0.1 \ kg$,$v = 10 \ m/s$.
$\lambda = \frac{6.626 \times 10^{-34}}{0.1 \times 10} = \frac{6.626 \times 10^{-34}}{1} = 6.626 \times 10^{-34} \ m$.
This value is extremely small,which indicates that the wave nature of macroscopic objects is negligible and cannot be detected.

(N/A) The wavelength of an electron is calculated using the de Broglie relation $\lambda = \frac{h}{mv}$.
This wavelength is relatively long compared to the wavelength of matter and indicates that the electron possesses wave-particle duality,specifically exhibiting wave nature.

(B) Energy incident $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9}} \, J$
Converting to $eV$: $E = \frac{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}{248 \times 10^{-9} \times 1.6 \times 10^{-19}} \approx 5.0 \, eV$
Using Einstein's photoelectric equation: $E = \phi + K.E.$
$5.0 \, eV = 3.0 \, eV + K.E. \implies K.E. = 2.0 \, eV$
$K.E. = 2.0 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-19} \, J$
De-Broglie wavelength $\lambda = \frac{h}{\sqrt{2mK.E.}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{58.24 \times 10^{-50}}} = \frac{6.63 \times 10^{-34}}{7.63 \times 10^{-25}} \approx 0.868 \times 10^{-9} \, m$
$\lambda \approx 8.68 \times 10^{-10} \, m = 8.68 \, \mathring{A}$
Rounding to the nearest integer,we get $9 \, \mathring{A}$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For a proton $(P)$ and $Li^{3+}$ nucleus accelerated by the same potential $V$:
$\frac{\lambda_{Li}}{\lambda_{P}} = \sqrt{\frac{m_{P} \times q_{P}}{m_{Li} \times q_{Li}}} = \sqrt{\frac{m_{P} \times e}{8.3 m_{P} \times 3e}} = \sqrt{\frac{1}{8.3 \times 3}} = \sqrt{\frac{1}{24.9}} \approx \sqrt{0.04016} \approx 0.2004$.
Given $\frac{\lambda_{Li}}{\lambda_{P}} = x \times 10^{-1}$,we have $0.2004 = x \times 10^{-1}$,so $x \approx 2.004$.
Rounding to the nearest integer,$x = 2$.

(D) The de Broglie wavelength of an electron is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.
Since the electron is accelerated from rest,$KE = qV$.
Thus,$\lambda = \frac{h}{\sqrt{2mqV}}$.
Substituting the given values:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40 \times 10^{3}}}$.
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{116.48 \times 10^{-47}}} = \frac{6.63 \times 10^{-34}}{\sqrt{11.648 \times 10^{-46}}}$.
$\lambda = \frac{6.63 \times 10^{-34}}{3.413 \times 10^{-23}} \approx 1.94 \times 10^{-11} \, m$ (Wait,re-calculating).
Using the shortcut formula: $\lambda = \frac{12.27}{\sqrt{V}} \, \mathring{A} = \frac{12.27}{\sqrt{40000}} \, \mathring{A} = \frac{12.27}{200} \, \mathring{A} = 0.06135 \, \mathring{A}$.
Converting to meters: $0.06135 \times 10^{-10} \, m = 6.135 \times 10^{-12} \, m$.
Rounding to the nearest integer,$X = 6$.

(B) Given that the de Broglie wavelengths are equal: $\lambda_{e} = \lambda_{N}$.
Using the de Broglie relation $\lambda = \frac{h}{mv}$,we have $\frac{h}{m_{e} v_{e}} = \frac{h}{m_{N} v_{N}}$.
This simplifies to $m_{e} v_{e} = m_{N} v_{N}$,or $v_{e} = \frac{m_{N}}{m_{e}} v_{N}$.
Given $v_{e} = x \ v_{N}$,we have $x = \frac{m_{N}}{m_{e}}$.
Substituting the values: $x = \frac{1.6 \times 10^{-27} \ kg}{9.1 \times 10^{-31} \ kg}$.
$x = \frac{1.6}{9.1} \times 10^{4} = 0.175824 \times 10^{4} = 1758.24$.
The nearest integer value of $x$ is $1758$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{660 \times 10^{-9}} = 3.01 \times 10^{-19} \, J$.
The work function $\Phi = 1.0 \, eV = 1.6 \times 10^{-19} \, J$.
The kinetic energy of the ejected electron is $KE = E - \Phi = 3.01 \times 10^{-19} - 1.6 \times 10^{-19} = 1.41 \times 10^{-19} \, J$.
The de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2m KE}}$.
Substituting $m = 9.1 \times 10^{-31} \, kg$ and $h = 6.63 \times 10^{-34} \, J \cdot s$:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.41 \times 10^{-19}}} \approx 1.32 \times 10^{-9} \, m$.

(A) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
First,calculate the mass of one $C_{60}$ molecule in $kg$:
Mass of $C_{60} = 60 \times 12 \ \text{amu} = 720 \ \text{amu}$.
Since $1 \ \text{amu} = 1.66 \times 10^{-27} \ kg$,the mass $m = 720 \times 1.66 \times 10^{-27} \ kg \approx 1.195 \times 10^{-24} \ kg$.
Given $\lambda = 11.0 \ \mathring{A} = 11.0 \times 10^{-10} \ m$ and $h = 6.626 \times 10^{-34} \ J \ s$.
Rearranging for $v$: $v = \frac{h}{m \lambda} = \frac{6.626 \times 10^{-34}}{1.195 \times 10^{-24} \times 11.0 \times 10^{-10}}$.
$v \approx \frac{6.626 \times 10^{-34}}{1.3145 \times 10^{-33}} \approx 0.504 \ m \ s^{-1}$.
Comparing this with the given options,the value is closest to $0.5 \ m \ s^{-1}$,which corresponds to option $(A)$.

(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.
Given: Mass $m = 20 \, g = 20 \times 10^{-3} \, kg = 0.02 \, kg$.
Velocity $v = 100 \, ms^{-1}$.
Planck's constant $h = 6.626 \times 10^{-34} \, Js$.
Substituting the values into the formula:
$\lambda = \frac{6.626 \times 10^{-34} \, Js}{0.02 \, kg \times 100 \, ms^{-1}}$
$\lambda = \frac{6.626 \times 10^{-34}}{2} \, m$
$\lambda = 3.313 \times 10^{-34} \, m$.
Therefore,the correct option is $A$.

(D) The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.
Substituting the given values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times (9 \times 10^{-31} \ kg) \times (4.50 \times 10^{-29} \ J)}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{81 \times 10^{-60}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}}$
$\lambda = 0.733 \times 10^{-4} \ m = 7.33 \times 10^{-5} \ m$.
The nearest integer is $7$.

(B) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = a_0 n^2$.
For the $3^{rd}$ orbit $(n = 3)$, the radius is $r_3 = a_0 \times 3^2 = 9 a_0$.
According to Bohr's postulate, the circumference of the orbit is an integral multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.
Substituting the values for the $3^{rd}$ orbit: $2 \pi (9 a_0) = 3 \lambda$.
Solving for $\lambda$: $\lambda = \frac{18 \pi a_0}{3} = 6 \pi a_0$.

(A) According to de-Broglie's relation,the wavelength $(\lambda)$ is inversely proportional to the momentum $(p)$:
$\lambda = \frac{h}{p}$
where $h$ is Planck's constant.
This equation can be rewritten as $\lambda p = h$.
Since $h$ is a constant,the product of $\lambda$ and $p$ is constant.
This represents the equation of a rectangular hyperbola.
Therefore,the graph of $\lambda$ versus $p$ is a rectangular hyperbola.

(D) According to Bohr's quantization rule:
$2 \pi r_n = n \lambda$
Radius of $n^{th}$ orbit,$r_n = a_0 \frac{n^2}{Z}$
For hydrogen atom,$Z = 1$ and given $n = 4$.
$2 \pi \left(a_0 \frac{4^2}{1}\right) = 4 \lambda$
$2 \pi a_0 (16) = 4 \lambda$
$32 \pi a_0 = 4 \lambda$
$\lambda = 8 \pi a_0$
Therefore,the value is $8$.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
The frequency $\nu$ is given by $\nu = \frac{v}{\lambda}$.
Substituting $\lambda = \frac{h}{mv}$,we get $\nu = \frac{v \cdot mv}{h} = \frac{mv^2}{h}$.
From Bohr's model,the kinetic energy $K.E. = \frac{1}{2}mv^2 = 2.18 \times 10^{-18} \ J$.
Therefore,$mv^2 = 2 \times 2.18 \times 10^{-18} = 4.36 \times 10^{-18} \ J$.
Now,calculate the frequency: $\nu = \frac{4.36 \times 10^{-18}}{6.6 \times 10^{-34}} \ Hz$.
$\nu = 0.6606 \times 10^{16} \ Hz = 660.6 \times 10^{13} \ Hz$.
Rounding to the nearest integer,we get $661 \times 10^{13} \ Hz$.

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2m(KE)}}$.
Since $KE = \frac{3}{2}kT$,we have $\lambda \propto \frac{1}{\sqrt{mT}}$.
For $He$ at $-73^{\circ} C$ $(200 \ K)$: $m_{He} = 4$,$T_{He} = 200 \ K$.
For $Ne$ at $727^{\circ} C$ $(1000 \ K)$: $m_{Ne} = 20$,$T_{Ne} = 1000 \ K$.
$\frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{m_{Ne} T_{Ne}}{m_{He} T_{He}}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$.
Thus,$M = 5$.

(C) According to the Bohr quantization condition,the circumference of the orbit is an integral multiple of the de-Broglie wavelength:
$2 \pi r_n = n \lambda$
The radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$.
Substituting $r_n$ into the equation:
$2 \pi (n^2 a_0) = n \lambda$
Solving for $\lambda$:
$\lambda = \frac{2 \pi n^2 a_0}{n} = 2 \pi n a_0$
Note: The original question asked for the second orbit $(n=2)$,which would yield $\lambda = 4 \pi a_0$. Given the options provided,the expression for the $n^{th}$ orbit is derived as $2 \pi n a_0$.

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{\sqrt{2mK.E.}}$
Given:
$h = 6.6 \times 10^{-34} \ Js$
$m = 9 \times 10^{-31} \ kg$
$K.E. = 1.25 \times 10^{-28} \ J$
Substituting the values:
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 1.25 \times 10^{-28}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{22.5 \times 10^{-59}}} = \frac{6.6 \times 10^{-34}}{\sqrt{225 \times 10^{-60}}}$
$\lambda = \frac{6.6 \times 10^{-34}}{15 \times 10^{-30}} = 0.44 \times 10^{-4} \ m = 4.4 \times 10^{-5} \ m$

(B) The de-Broglie wavelength $(\lambda)$ is given by the formula: $\lambda = \frac{h}{mv}$
Where:
$h$ (Planck's constant) = $6.63 \times 10^{-34} \ J \cdot s$
$m$ (mass) = $10^{-6} \ kg$
$v$ (velocity) = $10 \ ms^{-1}$
Substituting the values:
$\lambda = \frac{6.63 \times 10^{-34}}{10^{-6} \times 10} = \frac{6.63 \times 10^{-34}}{10^{-5}} = 6.63 \times 10^{-29} \ m$
Therefore,the correct option is $B$.

(C) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For a proton,$\lambda_p = \frac{h}{\sqrt{2 m_p q_p V}}$.
For an $\alpha$-particle,the mass $m_{\alpha} = 4 m_p$ and the charge $q_{\alpha} = 2 q_p$.
Substituting these values,$\lambda_{\alpha} = \frac{h}{\sqrt{2 (4 m_p) (2 q_p) V}} = \frac{h}{\sqrt{16 m_p q_p V}} = \frac{1}{4} \frac{h}{\sqrt{m_p q_p V}}$.
Comparing this with $\lambda_p$,we get $\lambda_{\alpha} = \frac{\lambda}{2 \sqrt{2}}$.

(B) The energy of separation is given by $E_{\text{sep}} = \frac{13.6 \ Z^2}{n^2} \ eV$. For a Hydrogen atom,$Z = 1$. Given $E_{\text{sep}} = 3.4 \ eV$,we have $3.4 = \frac{13.6}{n^2}$,which gives $n^2 = 4$,so $n = 2$.
According to Bohr's postulate,the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $2 \pi r_n = n \lambda$.
The radius of the $n^{th}$ orbit is $r_n = r_1 \times n^2$,where $r_1 = 0.53 \ \mathring{A}$.
Substituting the values: $2 \pi \times (0.53 \times n^2) = n \lambda$.
Simplifying for $\lambda$: $\lambda = 2 \pi \times 0.53 \times n$.
Since $n = 2$,$\lambda = 2 \times 3.14159 \times 0.53 \times 2 \approx 6.66 \ \mathring{A}$.

(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$, where $a_0$ is the radius of the first orbit.
For the second orbit $(n = 2)$, the radius $r_2 = a_0 \times (2)^2 = 4 a_0$.
According to the Bohr quantization condition, the circumference of the orbit is an integral multiple of the de-Broglie wavelength: $n \lambda = 2 \pi r_n$.
For $n = 2$, we have $2 \lambda = 2 \pi r_2$.
Substituting $r_2 = 4 a_0$, we get $2 \lambda = 2 \pi (4 a_0)$.
Therefore, $\lambda = 4 \pi a_0$.

(A) According to the de Broglie relation,the momentum $P$ is given by $P = \frac{h}{\lambda}$.
Given,$\lambda = 1 \ \mathring{A} = 10^{-8} \ cm$ and $h = 6.6252 \times 10^{-27} \ erg \ s$.
Substituting the values,we get $P = \frac{6.6252 \times 10^{-27} \ erg \ s}{10^{-8} \ cm}$.
$P = 6.6252 \times 10^{-19} \ g \ cm / s$ (since $1 \ erg = 1 \ g \ cm^2 / s^2$,so $1 \ erg \ s / cm = 1 \ g \ cm / s$).