Question

The work function of a photosensitive metallic surface is \( h\nu_0 \). If photons of energy \( (2.5)h\nu_0 \) fall on this surface, the electrons come out with maximum velocity ' v '. When the photon energy is increased to \( 7h\nu_0 \), the maximum velocity of photoelectrons will be

• A. $2v$
• B. $2.5 \text{ v}$
• C. $3.5 \text{ v}$
• D. $4v$

Hint:

- KE $\propto$ $(\nu - \nu_0)$ - Velocity $\propto \sqrt{\text{KE}}$

Answer 1

The Correct Option is A

Concept: Photoelectric equation: \[ \frac{1}{2}mv^2 = h\nu - h\nu_0 \]

Step 1: Initial condition.
\[ \frac{1}{2}mv^2 = (2.5 - 1)h\nu_0 = 1.5h\nu_0 \]

Step 2: New condition.
\[ \frac{1}{2}mv'^2 = (7 - 1)h\nu_0 = 6h\nu_0 \]

Step 3: Take ratio.
\[ \frac{v'^2}{v^2} = \frac{6}{1.5} = 4 \Rightarrow \frac{v'}{v} = 2 \]