Question

The wavelength and intensity of the incident light is 4000 Å and 0.1 W respectively. What is the minimum change in the light energy? What is the number of incident photons? ​

Answer 1

Explanation:

We know that ,

Maximum energy of emitted photons electron

KE energy of the incident radiation - work function

=12.37eV

Now the kinetic energy is.....

KE= 12.37 - 2

= 10.37eV

Hence the kinetic energy is 10.37eV

Answer 2

Given,

The incident intensity =  0.1 W

The wavelength (λ) = 4000 $\AA$

The minimum change in the light energy will be given by:

$E=\frac{hc}{\lambda} = \frac{6.63\times 10^{-34}\times 3\times 10^{8} }{4000\times 10^{-10} } \\E= 4.97\times 10^{-19} J$

The energy is 4.97 × 10⁻¹⁹ J.

The minimum change is minor, hence, the change can be considered continuous.

The number of incident photons is:

$N=\frac{0.1W}{4.97\times10^{-19} } =2\times 10^{17}$

Hence, the number of incident photons is 2 × 10¹⁷.

The answer is 4.97 × 10⁻¹⁹ J and 2 × 10¹⁷ photons.