A photosensitive metallic surface has work function ϕ . If photon of energy 3 ϕ falls on the surface, the electron comes out with a maximum velocity of 6 × 1 0 6 m / s . When the photon energy is incr

Question

A photosensitive metallic surface has work function ϕ . If photon of energy 3 ϕ falls on the surface, the electron comes out with a maximum velocity of 6 × 1 0 6 m / s . When the photon energy is increased to 9 ϕ , then maximum velocity of photoelectrons will be

TestHub · Accepted Answer

According to Einstein's photoelectric equation:

$K_{m a x} = E - ϕ$

For the first case, the incident photon energy is $E_{1} = 3 ϕ$ . The maximum kinetic energy is:

$\frac{1}{2} m v_{1}^{2} = 3 ϕ - ϕ = 2 ϕ$

For the second case, the incident photon energy is $E_{2} = 9 ϕ$ . The maximum kinetic energy is:

$\frac{1}{2} m v_{2}^{2} = 9 ϕ - ϕ = 8 ϕ$

Taking the ratio of the kinetic energies in the two cases:

$\frac{\frac{1}{2} m v _{2}^{2}}{\frac{1}{2} m v _{1}^{2}} = \frac{8 ϕ}{2 ϕ}$

$\frac{v _{2}^{2}}{v _{1}^{2}} = 4$

$v_{2} = 2 v_{1}$

Given $v_{1} = 6 \times 1 0^{6} m/s$ , substituting this value gives:

$v_{2} = 2 \times 6 \times 1 0^{6} m/s = 12 \times 1 0^{6} m/s$

Answer: $12 \times 1 0^{6}, m / s$

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