Question

When a light of wavelength '$\lambda$' falls on the emitter of a photocell, the maximum speed of emitted photoelectrons is '$V$'. If the incident wavelength is changed to $\frac{2\lambda}{3}$, the maximum speed of emitted photoelectrons will be

• A. less than $V \left(1.5\right)^{1/2}$
• B. equal to $V$
• C. greater than $V \left(1.5\right)^{1/2}$
• D. equal to $V \left(1.5\right)^{1/2}$

Hint:

When the incident photon energy increases by a certain factor (here, wavelength drops to $\frac{2}{3}$ so energy multiplies by $1.5$), the work function stays fixed. Since the work function doesn't scale up, the entire excess energy is transferred directly into the electron's kinetic energy. This causes the maximum kinetic energy to increase by a factor even larger than $1.5$, making the final speed strictly greater than $\sqrt{1.5}V$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
In the photoelectric effect, wavelength λ produces max speed V. Find the inequality for new speed V' when wavelength becomes 2λ/3.

Step 2: Key Formula or Approach:
Einstein's photoelectric equation: hc/λ = φ₀ + ½mV². Decreasing wavelength increases photon energy, boosting the kinetic energy term.

Step 3: Detailed Explanation:
Case 1: hc/λ = φ₀ + ½mV². Case 2: hc/(2λ/3) = 1.5(hc/λ) = φ₀ + ½mV'². Substituting from Case 1: 1.5(φ₀ + ½mV²) = φ₀ + ½mV'² → ½mV'² = 1.5(½mV²) + 0.5φ₀. Since φ₀>0, V'²>1.5V², so V'>V(1.5)^(1/2).

Step 4: Final Answer:
The new speed satisfies V'>V(1.5)^(1/2), option (C).