Question

When a light of wavelength '\(λ\)' falls on the emitter of a photosensitive surface, maximum speed of emitted photoelectrons is 'V'. If the incident wavelength is changed to '\(2λ/3\)', maximum speed of emitted photoelectrons will be

• A. less than \(V(1.5)^{\frac{1}{2}}\)
• B. greater than \(V(1.5)^{\frac{1}{2}}\)
• C. less than \(V\)
• D. less than \(\frac{V}{2}\)

Hint:

Use Einstein photoelectric equation and keep the work function in the equation.

Answer 1

The Correct Option is B

Step 1: First case
\(\frac12mV^2 = \frac{hc}{\lambda} - \phi\).

Step 2: Second case
The new wavelength is \(\frac{2\lambda}3\), so the photon energy is \(\frac{3hc}{2\lambda} = 1.5\frac{hc}\lambda\). Then \(\frac12mV'^2 = 1.5\frac{hc}{\lambda} - \phi\).

Step 3: Compare
Substitute \(\frac{hc}\lambda = \frac12mV^2+\phi\): \(\frac12mV'^2 = 1.5\cdot\frac12mV^2 + 0.5\phi\). Because \(\phi>0\), \(V'^2 > 1.5V^2\).

Step 4: Result
So the speed is greater than \(V(1.5)^{1/2}\). Option (B).

Final Answer:
The new speed is greater than V times root 1.5. \[ \boxed{\text{(B)}\ \text{greater than }V(1.5)^{1/2}} \]