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: Understanding the Concept
Einstein's photoelectric equation is \(\dfrac{hc}{\lambda}=\phi_0+\dfrac12 mv^{2}\), where \(\phi_0\) is the work function.

Step 2: Write both cases
Case 1: \(\dfrac{hc}{\lambda}=\phi_0+\dfrac12 mv^{2}\).
Case 2: \(\dfrac{3hc}{\lambda}=\phi_0+\dfrac12 mv'^{2}\).

Step 3: Compare
Substituting from case 1: \(\dfrac12 mv'^{2}=3\left(\phi_0+\dfrac12 mv^{2}\right)-\phi_0=\dfrac32 mv^{2}+2\phi_0\).

Step 4: Conclusion
Since \(\phi_0>0\), \(v'^{2}>3v^{2}\), so \(v'>\sqrt{3}\,v\).

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