Question

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K.$ If $K$ is to be increased to exactly twice its original value, the wavelength must be changed to $\lambda$ such that

• A. ${\lambda }^{\prime }<\frac{\lambda }{2}$
• B. ${\lambda }^{\prime }>\frac{\lambda }{2}$
• C. $\lambda >{\lambda }^{\prime }>\frac{\lambda }{2}$
• D. ${\lambda }^{\prime }=\frac{\lambda }{2}$

Answer 1

The correct option is C $\lambda >{\lambda }^{\prime }>\frac{\lambda }{2}$

From, Einstein’s photoelectric equation, the energy of the incident photon is,

$E={\varphi }_0+KE$

$\Rightarrow \frac{hc}{\lambda }={\varphi }_0+KE.......\left(1\right)$

Case$-I$ :

$\Rightarrow \frac{hc}{\lambda }={\varphi }_0+K{E}_1........\left(2\right)$

Case$-II$ :

Now since the kinetic energy is doubled, thus equation (2) becomes,

$\Rightarrow \frac{hc}{{\lambda }^{\prime }}={\varphi }_0+2K{E}_1.......\left(3\right)$

From (1) and (2) we get,

$\Rightarrow \frac{\frac{hc}{\lambda }}{\frac{hc}{{\lambda }^{\prime }}}=\frac{{\varphi }_0+K{E}_1}{{\varphi }_0+2K{E}_1}$

$\Rightarrow \frac{{\lambda }^{\prime }}{\lambda }=\frac{{\varphi }_0+K{E}_1}{{\varphi }_0+2K{E}_1}$

$\therefore \frac{{\lambda }^{\prime }}{\lambda }<1$

Now we can clearly say that, ${\lambda }^{\prime }<\lambda$

Also, $\frac{{\lambda }^{\prime }}{\lambda }=\frac{1}{2}\left[\frac{K{E}_1+{\varphi }_0}{K{E}_1+\frac{{\varphi }_0}{2}}\right]$

$\frac{{\lambda }^{\prime }}{\lambda }>\frac{1}{2}$

${\lambda }^{\prime }>\frac{\lambda }{2}$

From the above conclusion, we can say

$\lambda >{\lambda }^{\prime }>\frac{\lambda }{2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.