Question

Energy required to remove an electron from an aluminium surface is 4.2 eV. If light of wavelength 2000 Å falls on the surface, the velocity of the fastest electron ejected from the surface will be

• A. \( 8.4 \times 10^5 \) ms\(^{-1}\)
• B. \( 7.4 \times 10^5 \) ms\(^{-1}\)
• C. \( 6.4 \times 10^5 \) ms\(^{-1}\)
• D. \( 8.4 \times 10^6 \) ms\(^{-1}\)

Hint:

To find the maximum velocity of an ejected electron, use Einstein’s photoelectric equation and the kinetic energy formula.

Answer 1

The Correct Option is A

Using Einstein’s photoelectric equation: \[ K_{\max} = h\nu - \phi \] where: - \( h = 6.63 \times 10^{-34} \) Js (Planck’s constant) - \( c = 3 \times 10^8 \) m/s (Speed of light) - \( \lambda = 2000 \) Å = \( 2 \times 10^{-7} \) m - \( \phi = 4.2 \) eV (Work function) First, find the energy of the incident photon: \[ E = \frac{hc}{\lambda} \] \[ = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{2 \times 10^{-7}} \] \[ = 9.945 \times 10^{-19} J \] Convert to eV: \[ E = \frac{9.945 \times 10^{-19}}{1.6 \times 10^{-19}} = 6.22 \text{ eV} \] Now, find the kinetic energy: \[ K_{\max} = 6.22 - 4.2 = 2.02 \text{ eV} \] Convert to joules: \[ K_{\max} = 2.02 \times 1.6 \times 10^{-19} = 3.23 \times 10^{-19} \text{ J} \] Using kinetic energy formula: \[ K = \frac{1}{2} m v^2 \] where \( m = 9.1 \times 10^{-31} \) kg (mass of electron), \[ v = \sqrt{\frac{2K}{m}} \] \[ = \sqrt{\frac{2 \times 3.23 \times 10^{-19}}{9.1 \times 10^{-31}}} \] \[ = \sqrt{7.1 \times 10^{11}} \] \[ = 8.4 \times 10^5 \text{ m/s} \] Thus, the correct answer is \( 8.4 \times 10^5 \) ms\(^{-1}\).