Question

Radiation of wavelength \(\lambda\), is incident on a photocell. The fastest emitted electron has speed \(v\). If the wavelength is changed to \(\frac{3\lambda }{4}\), the speed of the fastest electron will be

• A. greater than \(v\left(\frac{4}{3}\right)^{1/2}\)
• B. less than \(v\left(\frac{4}{3}\right)^{1/2}\)
• C. equal to \(v\left(\frac{4}{3}\right)^{1/2}\)
• D. equal to \(v\left(\frac{3}{4}\right)^{1/2}\)

Hint:

Compare the maximum kinetic energies for the two wavelengths; the work function is subtracted, so the ratio is not simply 4/3.

Answer 1

The Correct Option is A

Step 1: Energy ratio
If the work function were zero the kinetic energy would scale as $\dfrac{4}{3}$, giving $v'=v\sqrt{4/3}$.

Step 2: Effect of the work function
The kinetic energy is the photon energy minus $\phi$. Increasing the photon energy by a factor $\dfrac{4}{3}$ increases the difference by more than that factor, since $\dfrac{4}{3}(E-\phi)<\dfrac{4}{3}E-\phi$.

Step 3: Conclusion
So $K'>\dfrac{4}{3}K$ and the speed is greater than $v\left(\dfrac{4}{3}\right)^{1/2}$, option (A).

Final Answer:
The speed is greater than v times root of 4/3. \[ \boxed{\text{(A) }\text{greater than }v\left(\dfrac{4}{3}\right)^{1/2}} \]