Question

Light of wavelength \(\lambda\) strikes a photoelectric surface and electrons are ejected with energy \(E\). If \(E\) is to be increased to twice the original value, the wavelength changes to \(\lambda_1\)

• A. \(\lambda_1<\lambda/2\)
• B. \(\lambda_1 = \lambda\)
• C. \(\lambda_1>\lambda/2\)
• D. \(\lambda_1 = \lambda/2\)

Hint:

Use Einstein's photoelectric equation twice, once for the original wavelength and once when the kinetic energy is doubled.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Einstein's equation: \(E=\dfrac{hc}{\lambda}-\phi\), where \(\phi\) is the work function.

Step 2: Write both cases
Case 1: \(E=\dfrac{hc}{\lambda}-\phi\)
Case 2: \(2E=\dfrac{hc}{\lambda_1}-\phi\)

Step 3: Calculation
\(\dfrac{hc}{\lambda_1}=2E+\phi=2\left(\dfrac{hc}{\lambda}-\phi\right)+\phi=\dfrac{2hc}{\lambda}-\phi\)

Step 4: Compare the options
Since \(\phi\) is positive, \(\dfrac{hc}{\lambda_1}\lt\dfrac{2hc}{\lambda}\).
Hence \(\lambda_1\gt\dfrac{\lambda}{2}\), the third option.

Final Answer:
The new wavelength is greater than half of the original wavelength. \[ \boxed{\text{(C) }\lambda_1 \gt \lambda/2} \]