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Photoelectrons with a maximum speed of 8.00 × 107 m/s are ejected from a surface in the presence of light with a frequency of 7.50 × 1014Hz. If the mass of an electron is 9.10 × 10-31 kg, calculate the maximum kinetic energy of a single electron, in joules.


7.28 × 10-23 J

5.82 × 10-15 J

2.91 × 10-15 J

3.41 × 10-16 J

Verified Answer :

KE = 1/2 mass velocity^2
KE = 1/2 9.10x10^-31 (8.00x10^7)^2
KE = 1/2 9.10x10^-13 6.4x10^15
KE = 2.912x10^-15 J or C

Answer:

Maximum kinetic energy, \(E_k=2.91\times 10^{-15}\ J\)

Explanation:

It is given that,

Speed of photoelectrons, \(v_{max}=8\times 10^7\ m/s\)

Mass of electrons, \(m=9.1\times 10^{-31}\ Kg\)

Frequency of light, \(u=7.5\times 10^{14}\ Hz\)

According to Einstein equation :

\(h u=h u_o+\dfrac{1}{2}mv_{max}^2\)

Where

υ is frequency of radiation

υ₀ is the threshold frequency

Maximum kinetic energy of single electron is, \(E_k=\dfrac{1}{2}mv_{max}^2\)

\(E_k=\dfrac{1}{2}\times 9.1\times 10^{-31}\times (8\times 10^7)^2\)

\(E_k=2.91\times 10^{-15}\ J\)

Hence, the correct option is (C) " \(2.91\times 10^{-15}\ J\) "

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