Question

The kinetic energy of the electron is $E$ when the incident wavelength is $\lambda$ . To increase the kinetic energy of the electron to $2E$, the incident wavelength must be :

• A. 2$\lambda$
• B. $\frac{\lambda }{2}$
• C. $\frac{hc\lambda }{E\lambda +hc}$
• D. $\frac{2hc\lambda }{E\lambda +hc}$

Answer 1

The correct option is C $\frac{hc\lambda }{E\lambda +hc}$

Let Work function be $W$

According to Question,

$\frac{hc}{\lambda }-W=E\Rightarrow W=\frac{hc}{\lambda }-E=\frac{hc-E\lambda }{\lambda }$

So,

Now, let the new $\lambda$ be ${\lambda }_1$, to get K.E. as $2E$

So,

$\Rightarrow \frac{hc}{{\lambda }_1}-\frac{hc-E\lambda }{\lambda }=2E$

$\Rightarrow \frac{hc}{{\lambda }_1}=\frac{2E\lambda +hc-E\lambda }{\lambda }$

$\Rightarrow \frac{hc\lambda }{E\lambda +hc}={\lambda }_1$