Question

Threshold wavelength of a metal is 4000${A^0}$. If light of wavelength of wavelength 3000 ${A^0}$irradiates the surface, the maximum kinetic energy of the photoelectron is?
(A). 1.7 eV
(B). 1.6 eV
(C). 1.5 eV
(D). 1.0 eV

Answer 1

Hint: In this question, remember the concept and the equation of Einstein photoelectric equation use the equation and remember to change the unit of wavelength from ${A^0}$ to meter by multiplying by ${10^{-10}}$ using these steps will help you to get closer towards the solution to the questions.

Complete step-by-step solution -
According to the given information we have $\lambda = 3000{A^0}$ = 3000 $\times$ $10^{-10}$ = 3 $\times$ ${10^{-7}}$ m and${\lambda _0} = 4000{A^0}$= 4000 $\times$ ${10^{-10}}$ = 4 $\times$ ${10^{-7}}$ m, h = 6 $\times$ ${10^{-34}}$ Js and c = 3 $\times$ ${10^{8}}$ m/s
As we know that according to the Einstein photoelectric equation when a light of some sufficient frequency with some wavelength is projected on the surface of metal the electrons on the metal surface comes out of the surface this is effect is called the photoelectric effect
To calculate the energy of electron emitted from the metal surface Einstein gave an equation which is given as E = hv here h is planck constant and v is the frequency of light projected on the metal surface
So here according to Einstein when a photon is imparted on the metal surface it gives all the energy to the electron present on the metal surface which helps the electron to eject from the metal surface some energy is used to eject electron and the rest of energy act as the kinetic energy of electron
The above statement is given as Kinetic energy = energy of photon – Work function (energy required to eject electron) i.e. K.E = hv – W
Let’s find the maximum kinetic energy of photoelectron using the formula of kinetic energy
K.E = hv – W (equation 1)
As we know that $v = \dfrac{c}{\lambda }$ and W = $h{v_o}$
Substituting the given values in the equation 1
K.E = $\dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}$ (equation 2)
Now substituting the values given in the equation 2
K.E = $\dfrac{{6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3 \times {{10}^{ - 7}}}} - \dfrac{{6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4 \times {{10}^{ - 7}}}}$
$\Rightarrow$K.E = $\dfrac{{1.98 \times {{10}^{ - 19}}}}{{12 \times 1.6 \times {{10}^{ - 19}}}}$
$\Rightarrow$K.E = 1.03 eV $\approx$ 1 eV
So, the maximum kinetic energy of a photoelectron is 1 eV.

Hence option D is the correct option.

Note: As we discussed about the photoelectric effect in the above solution do you know what is threshold wavelength that we used in the above question many times the concept of threshold wavelength says that to release the electron form the metal surface the maximum wavelength or minimum frequency of a photon is required is called threshold wavelength under the photoelectric effect.