When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer:

the required wavelength of light is 161.9 nm

Explanation:

Given the data in the question;

Let us represent work function of the metal by W.

Now, using Einstein photoelectric effect equation;

[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]

hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1

we solve for W

W = hc/λ - [tex]K_{max[/tex]

given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J

we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.

so we substitute

W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] - 3.28 × 10⁻¹⁹

W = 8.99457 × 10⁻¹⁹ - 3.28 × 10⁻¹⁹

W = 5.71457 × 10⁻¹⁹ J

Now, to determine λ for which maximum kinetic energy is double

so;

[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J

from from equation 1

we solve for λ'

λ' = hc / W + [tex]K'_{max[/tex]

we substitute

λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))

λ' = 1.9878 × 10⁻²⁴ / 1.227457 × 10⁻¹⁸

λ' = 1.619 × 10⁻⁷ m

λ' = 161.9 nm

Therefore, the required wavelength of light is 161.9 nm