Question

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

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

Answer 1

Answer: (C)

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