Question

If \( \lambda_0 \) and \( \lambda \) are respectively the threshold wavelength and wavelength of incident light, the velocity of photoelectrons ejected from the metal surface is:

• A. \( \sqrt{\frac{2h}{m} (\lambda_0 - \lambda)} \)
• B. \( \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \)
• C. \( \sqrt{\frac{2hc}{m} (\lambda_0 - \lambda)} \)
• D. \( \sqrt{\frac{2h}{m} \left( \frac{1}{\lambda_0} - \frac{1}{\lambda} \right)} \)

Hint:

The velocity of emitted photoelectrons is given by \( v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \).

Answer 1

The Correct Option is B

Step 1: Apply Einstein’s Photoelectric Equation \[ \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \] Step 2: Compute Electron Velocity \[ v = \sqrt{\frac{2}{m} \left( \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \right)} \] \[ = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0 \lambda} \right)} \] Thus, the correct answer is \( \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \).

Answer 2

The photoelectric effect involves the ejection of electrons (photoelectrons) from a metal surface when light of sufficient energy falls on it.

Given:
- \( \lambda_0 \): threshold wavelength (minimum wavelength required to eject electrons)
- \( \lambda \): wavelength of the incident light
- \( h \): Planck's constant
- \( c \): speed of light
- \( m \): mass of the electron

Step 1: Energy of the incident photon:
\[ E = \frac{hc}{\lambda} \]

Step 2: Energy needed to eject an electron (work function \( \phi \)):
\[ \phi = \frac{hc}{\lambda_0} \]

Step 3: Kinetic energy (\( KE \)) of the ejected photoelectron:
\[ KE = E - \phi = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \]

Step 4: Expression for velocity (\( v \)) of photoelectron:
The kinetic energy is related to velocity as:
\[ KE = \frac{1}{2} m v^2 \] So, \[ v = \sqrt{\frac{2 KE}{m}} = \sqrt{\frac{2}{m} \cdot hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)} \]

Step 5: Simplify the expression inside the square root:
\[ \frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \] Assuming the incident wavelength \( \lambda \) is close to \( \lambda_0 \), the dominant term simplifies to:
\[ v = \sqrt{\frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right)} \]

Therefore, the velocity of photoelectrons ejected from the metal surface is:
\[ \boxed{ \sqrt{ \frac{2hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda_0} \right) } } \]