Question

An electron, practically at rest, is initially accelerated through a potential difference of $100\mathrm{V}$ . It then has a de Broglie wavelength $={\lambda }_1\mathrm{\AA }$. It then gets retarded through $19\mathrm{V}$ and then has wavelength ${\lambda }_2\mathrm{\AA }$. A further retardation through $32\mathrm{V}$ changes the wavelength to ${\lambda }_3\mathrm{\AA }$. What is $\frac{{\lambda }_3-{\lambda }_2}{{\lambda }_1}$

• A. $\frac{10}{63}$
• B. $\frac{20}{63}$
• C. $\frac{40}{63}$
• D. $\frac{1}{63}$

Answer 1

The correct option is B

$\frac{20}{63}$

Step 1: Given

Voltage the electron is subjected to initially, ${\mathrm{V}}_1=100\mathrm{V}$

Initial de Broglie wavelength is ${\lambda }_1$

The voltage gets retarded by $19\mathrm{V}$. Thus, the voltage to which the electron is subjected for the second time, ${\mathrm{V}}_2=100-19=81\mathrm{V}$

de Broglie wavelength for the second condition is ${\lambda }_2$

The voltage gets further retarded by $32\mathrm{V}$. Thus, the voltage to which the electron is subjected for the third time, ${\mathrm{V}}_3=81-32=49\mathrm{V}$

de Broglie wavelength for the second condition is ${\lambda }_3$

Step 2: Formulas used

De Broglie wavelength of a particle is given as,

$\lambda =\frac{\mathrm{h}}{\sqrt{2m_{e}eV}}$

Where, $\mathrm{h}$ is the planck's constant, $m_e$ is the mass of the electron and $V$ is the voltage, and $e$ is the charge of the electron

Step 3: Calculate ${\lambda }_1$

From de Broglie's equation for wavelength,

${\lambda }_1=\frac{\mathrm{h}}{\sqrt{2m_{e}e\times 100}}=\frac{\mathrm{h}}{\sqrt{200m_{e}e}}$

Step 4: Calculate ${\lambda }_2$

From de Broglie's equation for wavelength,

${\lambda }_2=\frac{\mathrm{h}}{\sqrt{2m_{e}e\times 81}}=\frac{\mathrm{h}}{\sqrt{162m_{e}e}}$

Step 5: Calculate ${\lambda }_3$

From de Broglie's equation for wavelength,

${\lambda }_3=\frac{\mathrm{h}}{\sqrt{2m_{e}e\times 49}}=\frac{\mathrm{h}}{\sqrt{98m_{e}e}}$

Step 6: Calculate the given ratio

Thus, the given ratio can be simplified as,

$\begin{array}{rcl}\frac{{\lambda }_3-{\lambda }_2}{{\lambda }_1}& =& \frac{{\displaystyle \frac{\mathrm{h}}{\sqrt{98m_{e}e}}}-{\displaystyle \frac{\mathrm{h}}{\sqrt{162m_{e}e}}}}{{\displaystyle \frac{\mathrm{h}}{\sqrt{200m_{e}e}}}}\\ & =& \frac{{\displaystyle \frac{\mathrm{h}}{\sqrt{m_{e}e}}}\left({\displaystyle \frac{1}{\sqrt{98}}}-{\displaystyle \frac{1}{\sqrt{162}}}\right)}{{\displaystyle \frac{\mathrm{h}}{\sqrt{m_{e}e}}}·{\displaystyle \frac{1}{\sqrt{200}}}}\\ & =& \frac{\sqrt{200}\left(\sqrt{162}-\sqrt{98}\right)}{\sqrt{98\times 162}}\\ & =& \frac{10\sqrt{2}\left(9\sqrt{2}-7\sqrt{2}\right)}{9\sqrt{2}\times 7\sqrt{2}}\\ \therefore \frac{{\lambda }_3-{\lambda }_2}{{\lambda }_1}& =& \frac{20}{63}\end{array}$

Therefore, $\frac{\begin{array}{rcl}& & {\lambda }_3-{\lambda }_2\end{array}}{\begin{array}{rcl}& & {\lambda }_1\end{array}}=\frac{20}{63}$.