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Question: The photoelectric threshold wavelength of silver is \[3250 \times {10^{ - 10}}m\]. The velocity of t...

The photoelectric threshold wavelength of silver is 3250×10−10m3250 \times {10^{ - 10}}m. The velocity of the electron ejected from a silver surface by the ultraviolet light of wavelength 2536×10−10m2536 \times {10^{ - 10}}mis:
(Given h=4.14×10−15eVsh = 4.14 \times {10^{ - 15}}eVs and c=3×108 m sec−1c = 3 \times {10^8}{\text{ m se}}{{\text{c}}^{ - 1}})
A. ≈0.3×106ms−1 \approx 0.3 \times {10^6}m{s^{ - 1}}
B. ≈6×105ms−1 \approx 6 \times {10^5}m{s^{ - 1}}
C. ≈6×106ms−1 \approx 6 \times {10^6}m{s^{ - 1}}
D. ≈61×103ms−1 \approx 61 \times {10^3}m{s^{ - 1}}

Explanation

Solution

Threshold frequency is the minimum frequency at which radiation produces a photoelectric effect. In the photoelectric effect, the light is incident on the metal surface to eject the electrons depending upon the frequency. An increase in the frequency of the incident light increases the kinetic energy of the emitted electrons; hence a greater retarding potential is required to stop the emission.
According to the Einstein theory, when a photon falls on the surface of a metal plate, the entire photon’s energy is transferred to the electron. This can be written as
The energy of Photon = Energy required to eject electron + Maximum kinetic energy

E=W+KE hν=W+KE KE=hν−KE  E = W + KE \\\ h\nu = W + KE \\\ KE = h\nu - KE \\\

The energy carried out by each particle of light is dependent on the light’s frequency (ν\nu ) is E=hνE = h\nu
The energy of a photon with frequencyν0{\nu _0} must be the work function w=hν0w = h{\nu _0}
Hence the maximum kinetic energy equation becomes KE=12mv2=hν−hν0KE = \dfrac{1}{2}m{v^2} = h\nu - h{\nu _0}

Complete step by step answer:
Given that the wavelengths of the silver and the ultraviolet light are:
λ=2536×10−10m=2536A∘\lambda = 2536 \times {10^{ - 10}}m = 2536{{\rm A}^ \circ }
λ0=3250×10−10m=3250A∘{\lambda _0} = 3250 \times {10^{ - 10}}m = 3250{{\rm A}^ \circ }
Using maximum kinetic energy equation

12mv2=hν−hν0 =hc[1λ−1λ0] [As,v=cλ]  \dfrac{1}{2}m{v^2} = h\nu - h{\nu _0} \\\ = hc\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right]{\text{ }}\left[ {As,v = \dfrac{c}{\lambda }} \right] \\\

This can be re-written as:

v2=2hcm[1λ−1λ0] v=2hcm[1λ−1λ0] =2×(4.14×10−15×1.6×10−19)×3×1089.1×10−31[12536×10−10−13250×10−10] =3.974×10−259.1×10−31[12536×10−10−13250×10−10] =2.72×1024[714×10−108242000×10−20] =4.3×105×8.66×105 =37.24×1010 =6.1×105m sec−1  {v^2} = \dfrac{{2hc}}{m}\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right] \\\ v = \sqrt {\dfrac{{2hc}}{m}\left[ {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _0}}}} \right]} \\\ = \sqrt {\dfrac{{2 \times \left( {4.14 \times {{10}^{ - 15}} \times 1.6 \times {{10}^{ - 19}}} \right) \times 3 \times {{10}^8}}}{{9.1 \times {{10}^{ - 31}}}}\left[ {\dfrac{1}{{2536 \times {{10}^{ - 10}}}} - \dfrac{1}{{3250 \times {{10}^{ - 10}}}}} \right]} \\\ = \sqrt {\dfrac{{3.974 \times {{10}^{ - 25}}}}{{9.1 \times {{10}^{ - 31}}}}\left[ {\dfrac{1}{{2536 \times {{10}^{ - 10}}}} - \dfrac{1}{{3250 \times {{10}^{ - 10}}}}} \right]} \\\ = \sqrt {2.72 \times {{10}^{24}}\left[ {\dfrac{{714 \times {{10}^{ - 10}}}}{{8242000 \times {{10}^{ - 20}}}}} \right]} \\\ = \sqrt {4.3 \times {{10}^5} \times 8.66 \times {{10}^5}} \\\ = \sqrt {37.24 \times {{10}^{10}}} \\\ = 6.1 \times {10^5}{\text{m se}}{{\text{c}}^{ - 1}} \\\

Hence the velocity of the electron ejected from a silver surface 6.1×105m sec−16.1 \times {10^5}{\text{m se}}{{\text{c}}^{ - 1}}

So, the correct answer is “Option B”.

Note:
Students must note that the threshold frequency depends on the nature of the material used of the emitted electron. If the frequency of the light is less than the threshold frequency, then even an increase in intensity does not cause the emission of an electron.