Question

If $\lambda_{0}$ and $\lambda$ be the threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is :

• A. $\sqrt{\frac{2h}{m}\left(\lambda_{0}-\lambda\right)}$
• B. $\sqrt{\frac{2hc}{m}\left(\lambda_{0}-\lambda\right)}$
• C. $\sqrt{\frac{2hc}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda\,\lambda_{0}}\right)}$
• D. $\sqrt{\frac{2h}{m}\left(\frac{1}{\lambda_{0}}-\frac{1}{\lambda}\right)}$

Answer 1

The Correct Option is C

To determine the velocity of the photoelectron ejected from the metal surface when light of wavelength $\lambda$ is incident, we must use the photoelectric effect equations. According to Einstein's photoelectric equation:

\[ K.E. = \frac{1}{2}mv^2 = h\nu - h\nu_{0} \]

where:

• $K.E.$ is the kinetic energy of the ejected photoelectron.
• $h$ is Planck's constant.
• $v$ is the velocity of the photoelectron.
• $\nu$ is the frequency of the incident light.
• $\nu_{0}$ is the threshold frequency.

By substituting frequency with wavelength (since $\nu = \frac{c}{\lambda}$), we can rewrite the equation as:

\[ \frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} \]

Simplifying further:

\[ \frac{1}{2}mv^2 = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}\right) \]

To solve for the velocity $v$, rearrange and solve the equation:

\[ v = \sqrt{\frac{2hc}{m}\left(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}\right)} \]

Recognizing that:

\[ \frac{1}{\lambda} - \frac{1}{\lambda_{0}} = \frac{\lambda_{0} - \lambda}{\lambda \cdot \lambda_{0}} \]

Substitute back into the velocity equation:

\[ v = \sqrt{\frac{2hc}{m}\left(\frac{\lambda_{0} - \lambda}{\lambda \cdot \lambda_{0}}\right)} \]

This matches the provided correct answer option:

• $\sqrt{\frac{2hc}{m}\left(\frac{\lambda_{0}-\lambda}{\lambda\,\lambda_{0}}\right)}$

This equation provides the relationship between the velocity of the photoelectron and the wavelengths, building upon the classical understanding of the photoelectric effect.