Question

The work function of a metal is 1eV. Light of wavelength 3000Å is incident on this metal surface. The velocity of emitted photoelectrons will be

• A. 10 ms^-1
• B. 1 × 10³ ms^-1
• C. 1 × 10⁴ ms^-1
• D. 1 × 10⁶ ms^-1

Answer 1

The Correct Option is D

Given:

• Work function (Φ) = 1 eV = 1.6 × 10⁻¹⁹ J
• Wavelength of light (λ) = 3000 Å = 3000 × 10⁻¹⁰ m = 3 × 10⁻⁷ m
• Planck's constant (h) = 6.63 × 10⁻³⁴ Js
• Speed of light (c) = 3 × 10⁸ m/s
• Electron mass (mₑ) = 9.11 × 10⁻³¹ kg

Step 1: Calculate photon energy (E)

Using Planck's equation:

\[ E = \frac{hc}{λ} \]

\[ E = \frac{(6.63 × 10⁻³⁴)(3 × 10⁸)}{3 × 10⁻⁷} = 6.63 × 10⁻¹⁹ J \]

Convert to eV:

\[ E = \frac{6.63 × 10⁻¹⁹}{1.6 × 10⁻¹⁹} ≈ 4.14 eV \]

Step 2: Determine kinetic energy of photoelectrons

Using Einstein's photoelectric equation:

\[ E = Φ + K.E. \]

\[ K.E. = E - Φ = 4.14 eV - 1 eV = 3.14 eV \]

Convert to joules:

\[ K.E. = 3.14 × 1.6 × 10⁻¹⁹ = 5.02 × 10⁻¹⁹ J \]

Step 3: Calculate electron velocity

Using kinetic energy formula:

\[ K.E. = \frac{1}{2}mₑv² \]

\[ v = \sqrt{\frac{2 × K.E.}{mₑ}} \]

\[ v = \sqrt{\frac{2 × 5.02 × 10⁻¹⁹}{9.11 × 10⁻³¹}} \]

\[ v = \sqrt{1.10 × 10¹²} \]

\[ v ≈ 1.05 × 10⁶ m/s \]