Question

Initially a photon of wavelength ${\lambda }_1$ falls on photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to ${\lambda }_2$, the maximum energy of the electron emitted becomes $E_2$. Then value of hc (h = Planck's constant, c = Velocity of light) is :

• A. $hc=\frac{(E_1-E_2){\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$
• B. $hc=\frac{E_1+E_2}{{\lambda }_2-{\lambda }_1}\cdot \left({\lambda }_1{\lambda }_2\right)$
• C. $hc=\frac{(E_1+E_2)({\lambda }_2-{\lambda }_1)}{{\lambda }_1{\lambda }_2}$
• D. $hc=\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1{\lambda }_2{E}_1}\cdot E_1$

Answer 1

The correct option is A $hc=\frac{(E_1-E_2){\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$

We know that electron with maximum energy has an energy of $E=\frac{hc}{\lambda }-\varphi$ where $\varphi$ is work function of metal.

Hence, as per the problem,

$E_1=\frac{hc}{{\lambda }_1}-\varphi$ and $E_2=\frac{hc}{{\lambda }_2}-\varphi$.

Subtracting the two equations, we get,

$E_1-E_2=hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

$hc={\lambda }_1{\lambda }_2\times \frac{(E_1-E_2)}{{\lambda }_2-{\lambda }_1}$.