De Broglie's principle Questions in English

(C) The quantum mechanical model of an atom is based on the dual nature of matter (wave-particle duality) proposed by $de \text{ } Broglie$.
This model considers the electron as both a particle and a wave,which is mathematically described by the $Schrodinger$ wave equation.

(A) From de-Broglie's equation,$\lambda = \frac{h}{mv}$.
Squaring both sides,$\lambda^2 = \frac{h^2}{m^2v^2}$.
Rearranging for kinetic energy $(KE = \frac{1}{2}mv^2)$,we get $mv^2 = \frac{h^2}{m\lambda^2}$.
Therefore,$KE = \frac{1}{2} \times \frac{h^2}{m\lambda^2}$.
This implies $KE \propto \frac{1}{\lambda^2}$.
Given the ratio of wavelengths $\frac{\lambda_1}{\lambda_2} = \frac{3}{5}$,the ratio of kinetic energies is $\frac{K_1}{K_2} = \left(\frac{\lambda_2}{\lambda_1}\right)^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$.
Thus,the ratio is $25: 9$.

(A) Given,velocity of particle $A$ $(v_A)$ = $0.05 \ ms^{-1}$.
Velocity of particle $B$ $(v_B)$ = $0.02 \ ms^{-1}$.
Let the mass of particle $A$ $(m_A)$ = $m$.
Then,the mass of particle $B$ $(m_B)$ = $5m$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For particle $A$,$\lambda_A = \frac{h}{m_A v_A} = \frac{h}{m \times 0.05}$.
For particle $B$,$\lambda_B = \frac{h}{m_B v_B} = \frac{h}{5m \times 0.02} = \frac{h}{0.1m}$.
Taking the ratio $\frac{\lambda_A}{\lambda_B} = \frac{h}{m \times 0.05} \times \frac{0.1m}{h} = \frac{0.1}{0.05} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Given: $m_1 = 100 \ g$,$m_2 = 50 \ g$,and $v_2 = 1.5 \ v_1$.
The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h / (m_1 v_1)}{h / (m_2 v_2)} = \frac{m_2 v_2}{m_1 v_1}$.
Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{50 \times 1.5 \ v_1}{100 \times v_1} = \frac{75}{100} = \frac{3}{4}$.
Thus,the ratio is $3 : 4$.

(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.
Since the speed $(v)$ is the same for all particles,the wavelength is inversely proportional to the mass: $\lambda \propto \frac{1}{m}$.
The masses of the particles are in the order $m_\alpha > m_p > m_e$.
Therefore,the wavelength order is $\lambda_e > \lambda_p > \lambda_\alpha$.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m \cdot v}$.
For a gas,the most probable velocity is $C_{mp} = \sqrt{\frac{2KT}{m}}$.
Thus,$\lambda = \frac{h}{m \sqrt{\frac{2KT}{m}}} = \frac{h}{\sqrt{2mKT}}$.
Given $T_{He} = -73 + 273 = 200 \ K$ and $T_{Ne} = 727 + 273 = 1000 \ K$.
$\lambda_{He} = M \times \lambda_{Ne}$ $\Rightarrow \frac{h}{\sqrt{2 \times 4 \times K \times 200}} = M \times \frac{h}{\sqrt{2 \times 20 \times K \times 1000}}$.
$\frac{1}{\sqrt{1600K}} = M \times \frac{1}{\sqrt{40000K}}$.
$\frac{1}{40\sqrt{K}} = M \times \frac{1}{200\sqrt{K}}$.
$M = \frac{200}{40} = 5$.

(C) According to de-Broglie's wavelength formula,$\lambda = \frac{h}{mv}$.
Given,mass $m = 100 \ g$ and velocity $v = 100 \ cm \ s^{-1}$.
Planck's constant $h$ in $CGS$ units is $6.626 \times 10^{-27} \ erg \ s$ (approximated as $6.6 \times 10^{-27} \ erg \ s$).
Substituting the values into the formula:
$\lambda = \frac{6.6 \times 10^{-27} \ erg \ s}{100 \ g \times 100 \ cm \ s^{-1}}$
$\lambda = \frac{6.6 \times 10^{-27}}{10^4} \ cm$
$\lambda = 6.6 \times 10^{-31} \ cm$.