Question

In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:

Given E (in eV) \(=\frac{{1237}}{{\lambda {\rm{\;}}\left( {in{\rm{\;}}nm} \right)}}\)
1. 1.5 eV
2. 3.0 eV
3. 4.5 eV
4. 15.1 eV

Answer 1

Correct Answer - Option 1 : 1.5 eV

Concept:

Kinetic energy of the electron:

The maximum kinetic energy of the electron is the energy of the photon minus the threshold energy. This threshold energy we call the "work function" and we give it the symbol Φ

The maximum kinetic energy of the electrons is given by the photoelectric effect as,

\(K{E_{max}} = E--{\phi _0}\) 

(where E = energy of incident light and  \({\phi _0} = {\rm{work\;function}}\))

Work function:

Work function is defined as the minimum amount of energy required by an electron to escape from a metal surface.

Work function formula is given by,

\({\phi _0} = \frac{{{\rm{hc}}}}{{{\lambda _0}}}\)   (Here, h is the Planck’s constant)

Calculation:

Given,

Threshold wavelength of light, λ₀ = 380 nm

Wavelength of the incident light, λ = 260 nm

Work function is,

\({\phi _0} = \frac{{{\rm{hc}}}}{{{\lambda _0}}} = \;\frac{{1237}}{{380}}\;\) 

Here, h is the Planck’s constant

Energy of incident light is,

\(E = \frac{{{\rm{hc}}}}{\lambda } = \frac{{1237}}{{260}}\)  (Given)

Thus, the maximum Kinetic Energy of the electron is,

KE_max = E – ϕ₀

\({\rm{K}}{{\rm{E}}_{{\rm{max}}}} = \frac{{1237}}{{260}} - \frac{{1237}}{{380}}\)

\({\rm{K}}{{\rm{E}}_{{\rm{max}}}} = 1237\left[ {\frac{1}{{260}} - \frac{1}{{380}}} \right]\)

\({\rm{K}}{{\rm{E}}_{{\rm{max}}}} = 1237\left[ {\frac{{19 - 13}}{{4940}}} \right]\)

\({\rm{K}}{{\rm{E}}_{{\rm{max}}}} = 1237\left[ {\frac{6}{{4940}}} \right] = \frac{{7422}}{{4940}}\)

KE_max = 1.5 eV

Therefore, the maximum kinetic energy of emitted electrons will be 1.5 eV.