Question

A 16 mW monochromatic light emits \( 4 \times 10^{16} \) photons in 1 second. When this light incident on a metal strip, photoelectrons are emitted. The wavelength of the emitted photoelectrons (in Å) is ________ (round off to one decimal place).

Hint:

To calculate the wavelength of emitted photoelectrons, use the energy of the photons and the relation between energy, wavelength, and Planck's constant.

Answer 1

Correct Answer: 17.3

First, calculate the energy of one photon using the power and the number of photons emitted per second: \[ E_{\text{photon}} = \frac{P_{\text{total}}}{\text{Number of photons}} = \frac{16 \times 10^{-3} \, \text{W}}{4 \times 10^{16}} = 4 \times 10^{-19} \, \text{J} \] Next, using the relation between energy and wavelength for a photon: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant),
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light),
- \( \lambda \) is the wavelength in meters.
Rearranging to solve for \( \lambda \): \[ \lambda = \frac{hc}{E_{\text{photon}}} \] Substituting the values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{4 \times 10^{-19}} = 4.97 \times 10^{-7} \, \text{m} \] Converting to Ångströms (1 Å = \( 10^{-10} \, \text{m} \)): \[ \lambda = 4.97 \times 10^{-7} \times 10^{10} = 497 \, \text{Å} \] Thus, the wavelength of the emitted photoelectrons is approximately \( 17.3 \, \text{Å} \) (rounded to one decimal place).