Modern Physics

The maximum kinetic energy of emitted photoelectrons in the photoelectric effect is determined by the equation:

$%7BK_%7B%5Ctext%7Bmax%7D%7D%7D%3Dh%5Cnu-%5Cphi$

Where:

$h%20%3D$ Planck's constant ( $6.626%20%5Ctimes%2010%5E%7B-34%7D%5C%2C%5Ctext%7BJs%7D$ )
$%5Cnu%20%3D$ Frequency of the incident light
$%5Cphi%20%3D$ Work function of the metal

The frequency $%5Cnu$ can be related to the wavelength $%5Clambda$ using:

$%5Cnu%3D%5Cdfrac%7Bc%7D%7B%5Clambda%7D$

Given $%5Clambda%3D400%5C%2C%5Ctext%7Bnm%7D%3D400%5Ctimes10%5E%7B-9%7D%5C%2C%5Ctext%7Bm%7D$ , and the speed of light $c%3D3%5Ctimes10%5E8%5C%2C%5Ctext%7Bm%2Fs%7D$ , we find $%5Cnu$ as:

$%5Cnu%3D%5Cdfrac%7B3%5Ctimes10%5E8%7D%7B400%5Ctimes10%5E%7B-9%7D%7D%3D7.5%5Ctimes10%5E%7B14%7D%5C%2C%5Ctext%7BHz%7D$

The energy of the photons is obtained by:

$E_%7B%5Ctext%7Bphoton%7D%7D%3Dh%5Cnu$

Substituting the values:

$E_%7B%5Ctext%7Bphoton%7D%7D%3D6.626%5Ctimes10%5E%7B-34%7D%5Ctimes7.5%5Ctimes10%5E%7B14%7D%5Capprox4.97%5Ctimes10%5E%7B-19%7D%5C%2C%5Ctext%7BJ%7D$

Convert $E_%7B%5Ctext%7Bphoton%7D%7D$ to electronvolts (1 eV = $1.602%5Ctimes10%5E%7B-19%7D%5C%2C%5Ctext%7BJ%7D$ ):

$%5Capprox%20%5Cdfrac%7B4.97%5Ctimes10%5E%7B-19%7D%7D%7B1.602%5Ctimes10%5E%7B-19%7D%7D%5Capprox3.1%5C%2C%5Ctext%7BeV%7D$

The maximum kinetic energy $%7BK_%7B%5Ctext%7Bmax%7D%7D%7D$ is:

$%7BK_%7B%5Ctext%7Bmax%7D%7D%7D%3D3.1-%5Cphi%3D3.1-2.0%3D1.1%5C%2C%5Ctext%7BeV%7D$

Doubling the intensity of light increases the number of photons hitting the metal per second, but it does not change the energy per photon or the frequency of light. Therefore, the maximum kinetic energy of the emitted photoelectrons remains unchanged.

The correct answer is:

It remains the same.

High-Yield Trend

Chapter Questions
2 MCQs

Official Solution

Official Solution

Modern Physics

High-Yield Trend

Chapter Questions 2 MCQs

Official Solution

Official Solution

Chapter Questions
2 MCQs