Question

Photoelectric emission is observed from a metallic surface for frequencies ν₁ and ν₂ of the incident light rays (ν₁ > ν₂). If the ratio of the maximum value of the kinetic energy of the photoelectrons emitted in the first case to that in the second case is 2 : K, then the threshold frequency of the metallic surface is ______.

पर्याय

• `(Kν_1 - ν_2)/(K - 1)`

• `(Kν_1 - 2ν_2)/(K - 2)`

• `(K - 1)/(Kν_1 - ν_2)`

• `(K - 2)/(Kν_1 - 2ν_2)`

Answer 1

Photoelectric emission is observed from a metallic surface for frequencies ν₁ and ν₂ of the incident light rays (ν₁ > ν₂). If the ratio of the maximum value of the kinetic energy of the photoelectrons emitted in the first case to that in the second case is 2 : K, then the threshold frequency of the metallic surface is `underlinebb((Kν_1 - 2ν_2)/(K - 2))`.

Explanation:

From Einstein's equation,

`KE_{max} = hν - hν_0`

⇒ `KE_{max1} = hν_1 - hν_0` .......(i)

and `KE_{max2} = hν_2 - hν_0` .......(ii)

Dividing Eq. (i) by Eq. (ii), we get

`(KE_{max1})/(KE_{max2}) = (h(ν_1 - ν_0))/(h(ν_2 - ν_0))`

`2/K = (ν_1 - ν_0)/(ν_2 - ν_0)`

⇒ `2ν_2 - 2ν_0 = Kν_1 - Kν_0 ⇒ (K - 2)ν_0 = Kν_1 - 2ν_2`

or `ν_0 = (Kν_1 - 2ν_2)/(K - 2)`