Initially a photon of wavelength $\lambda_1$ falls on photocathode and emits an electron of maximum energy…

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 $h c$ ( $h=$ Planck's constant, $c=$ velocity of light) is
  1. $h c=\frac{\left(E_1+E_2\right) \lambda_1 \lambda_2}{\lambda_2-\lambda_1}$
  2. $h c=\frac{E_1-E_2}{\lambda_2-\lambda_1} \cdot\left(\lambda_1 \lambda_2\right)$
  3. $h c=\frac{\left(E_1-E_2\right)\left(\lambda_2-\lambda_1\right)}{\lambda_1 \lambda_2}$
  4. $h c=\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2 E_2} \cdot E_1$

Solution

From equation of photoelectric effect, we have $ \begin{aligned} & E_1=\frac{h c}{\lambda_1}-W \\ & E_2=\frac{h c}{\lambda_2}-W \end{aligned} $ where, $W$ is work function. $ \begin{aligned} & E_1+W=\frac{h c}{\lambda_1} \\ & E_2+W=\frac{h c}{\lambda_2} \end{aligned} $ From Eq. (iv) $ W=\frac{h c}{\lambda_2}-E_2, $ $\therefore$ Putting this value in Eq. (iii), we have $ \begin{aligned} & E_1+\frac{h c}{\lambda_1}-E_2-\frac{h c}{\lambda_1} \\ \Rightarrow \quad & E_1-E_2=h c\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) \\ \Rightarrow \quad & E_1-E_2=h c\left(\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right) \\ \Rightarrow \quad h c= & \frac{\left(E_1-E_2\right) \lambda_1 \lambda_2}{\left(\lambda_2-\lambda_1\right)} \end{aligned} $

Asked in: AP EAMCET 2013

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