Solveeit Logo
Login

Question

Question: An electromagnetic radiation of frequency \(3 \times {10^{15}}\) Hz falls on a photo surface whose w...

An electromagnetic radiation of frequency 3×10153 \times {10^{15}} Hz falls on a photo surface whose work function is 4 eV. Then the maximum velocity of the photoelectrons emitted from the surface is
(given h=6.6×10−34Js,m=9×10−31Kgh = 6.6 \times {10^{ - 34}}Js,m = 9 \times {10^{ - 31}}Kg )
A. 1.7×106ms−11.7 \times {10^6}m{s^{ - 1}}
B. 3.4×106ms−13.4 \times {10^6}m{s^{ - 1}}
C. 2.5×106ms−12.5 \times {10^6}m{s^{ - 1}}
D. 2.0×106ms−12.0 \times {10^6}m{s^{ - 1}}

Explanation

Solution

When light (electromagnetic radiation) falls on a surface, some energy is required to release photons from the surface and the rest is used to increase the K.E. of the ejected photons. According to photoelectric effect, K.E=E+Eo=hυ−hυoK.E = E + {E_o} = h\upsilon - h{\upsilon _o} , where hυoh{\upsilon _o} is the energy of the incident photons (Work function), hυh\upsilon is the energy of ejected photons. Substitute the values and find the value of K.E. Kinetic energy is also given as 12mv2\dfrac{1}{2}m{v^2} , where mass is given. Substitute the values of mass and obtained K.E. and find the value of the velocity v, where v is the velocity of the emitted photons.

Complete step by step answer:
We are given an electromagnetic radiation of frequency 3×10153 \times {10^{15}} Hz falls on a photo surface whose work function is 4 eV.
And we have to calculate the maximum velocity of the photoelectrons emitted from the surface.
K.E=E+Eo=hυ−hυoK.E = E + {E_o} = h\upsilon - h{\upsilon _o}, where h is the Planck’s constant, EE is the energy of the ejected photons from the photo surface and Eo{E_o} is the energy of the incident photons to the photo surface which can also be called as Wok function.
hυo=W K.E=hυ−W W=4eV,h=6.6×10−34Js,υ=3×1015Hz K.E=(6.6×10−34×3×1015)−4eV 1eV=1.6×10−19J   ⟹  K.E=(6.6×10−34×3×1015)−(4×1.6×10−19)   ⟹  K.E=(19.8×10−19)−(6.4×10−19)   ⟹  K.E=13.4×10−19=1.34×10−18J  h{\upsilon _o} = W \\\ K.E = h\upsilon - W \\\ W = 4eV,h = 6.6 \times {10^{ - 34}}Js,\upsilon = 3 \times {10^{15}}Hz \\\ K.E = \left( {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^{15}}} \right) - 4eV \\\ 1eV = 1.6 \times {10^{ - 19}}J \\\ \implies K.E = \left( {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^{15}}} \right) - \left( {4 \times 1.6 \times {{10}^{ - 19}}} \right) \\\ \implies K.E = \left( {19.8 \times {{10}^{ - 19}}} \right) - \left( {6.4 \times {{10}^{ - 19}}} \right) \\\ \implies K.E = 13.4 \times {10^{ - 19}} = 1.34 \times {10^{ - 18}}J \\\
Kinetic energy is also equal to 12mv2\dfrac{1}{2}m{v^2}
K.E=12mv2 v=2K.Em m=9×10−31Kg,K.E=1.34×10−18J v=2×1.34×10−189×10−31   ⟹  v=2.97×1012 ∴v=1.7×106m/s  K.E = \dfrac{1}{2}m{v^2} \\\ v = \sqrt {\dfrac{{2K.E}}{m}} \\\ m = 9 \times {10^{ - 31}}Kg,K.E = 1.34 \times {10^{ - 18}}J \\\ v = \sqrt {\dfrac{{2 \times 1.34 \times {{10}^{ - 18}}}}{{9 \times {{10}^{ - 31}}}}} \\\ \implies v = \sqrt {2.97 \times {{10}^{12}}} \\\ \therefore v = 1.7 \times {10^6}m/s \\\
The maximum velocity of the photoelectrons emitted from the surface is 1.7×106ms−11.7 \times {10^6}m{s^{ - 1}}

So, the correct answer is “Option A”.

Note:
The above given question describes a phenomenon called photoelectric effect, when light hits a surface, it ejects electrons from them and the photon delivers its total energy to a single electron; whereas Compton Effect is another phenomenon, the photon transfers part of its energy to a single electron. The definitions of these two phenomena are almost the same but they both are completely different. So be careful while defining photoelectric effects.