Question

In a photoelectric cell, for the incident radiation of wavelength \(\lambda\), the fastest electron has speed 'v'. If the wavelength of incident radiation is changed to \(\frac{\lambda }{3}\), the speed of the fastest emitted electron is

• A. \(v\left(\frac{1}{\sqrt{3}}\right)\)
• B. \(v(\sqrt{3})\)
• C. less than \(v(\sqrt{3})\)
• D. greater than \(v(\sqrt{3})\)

Hint:

Compare the kinetic energy from Einstein's equation before and after; the work function means the new energy exceeds three times the old one.

Answer 1

The Correct Option is D

Step 1: Formula-first approach
Maximum kinetic energy is $K=\dfrac{hc}{\lambda}-\phi_0$. The photon energy triples when the wavelength becomes $\dfrac{\lambda}{3}$.

Step 2: Compare the options
If there were no work function, $K$ would become exactly 3 times and the speed would be $v\sqrt{3}$. Options (A) and (B) assume this idea and ignore $\phi_0$.

Step 3: Effect of the work function
The work function is subtracted only once, so $K'=3K+2\phi_0>3K$. The speed is therefore greater than $v\sqrt{3}$, which is option (D).

Final Answer:
The speed is greater than $v\sqrt{3}$. \[ \boxed{\text{(D) }\text{greater than } v(\sqrt{3})} \]