Question

The light of wavelength '$\lambda$' is incident on the surface of metal of work function $\phi$ and emits the electron. The maximum velocity of electron emitted is [$m = \text{mass of electron and } h = \text{Planck's constant, } c = \text{velocity of light}$]

• A. $\left[ \frac{2(hc - \lambda)}{m\lambda} \right]^{\frac{1}{2}}$
• B. $\left[ \frac{2(hc - \phi)\lambda}{mc} \right]$
• C. $\left[ \frac{2(hc - \lambda)}{m\lambda} \right]$
• D. $\left[ \frac{2(hc - \lambda\phi)}{m\lambda} \right]^{\frac{1}{2}}$

Hint:

To quickly verify your algebraic steps, check the dimensions or units of the terms inside the parentheses. Since $hc/\lambda$ represents energy, the product $\lambda\phi$ ensures that both terms in the numerator have matching units of (Energy $\times$ Length), which makes the expression dimensionally correct.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Monochromatic light of wavelength λ strikes a metal with work function φ; express the maximum photoelectron velocity v_max.

Step 2: Key Formula or Approach:
Einstein's equation: ½mv_max² = hc/λ – φ. Solve for v_max.

Step 3: Detailed Explanation:
½mv_max² = (hc – λφ)/λ → v_max² = 2(hc – λφ)/(mλ) → v_max = [2(hc – λφ)/(mλ)]^(½).

Step 4: Final Answer:
v_max = [2(hc – λφ)/(mλ)]^(½), matching option (D).