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: Understanding the Concept
The maximum kinetic energy is \(K=\dfrac{hc}{\lambda}-\phi\) and \(K=\dfrac{1}{2}mv^{2}\), so \(v^{2}\propto K\).

Step 2: Write the two cases
Initially \(\dfrac{1}{2}mv^{2}=\dfrac{hc}{\lambda}-\phi\).
Finally \(\dfrac{1}{2}mv'^{2}=\dfrac{4hc}{3\lambda}-\phi\).

Step 3: Relate the two
Using \(\dfrac{hc}{\lambda}=\dfrac{1}{2}mv^{2}+\phi\): \(\dfrac{1}{2}mv'^{2}=\dfrac{4}{3}\left(\dfrac{1}{2}mv^{2}+\phi\right)-\phi=\dfrac{4}{3}\cdot\dfrac{1}{2}mv^{2}+\dfrac{\phi}{3}\).

Step 4: Compare the options
Since \(\phi>0\), \(v'^{2}>\dfrac{4}{3}v^{2}\), so \(v'>v\left(\dfrac{4}{3}\right)^{1/2}\), option (A).

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