The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is

Question

The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is (A) Double the earlier value (B) Unchanged (C) More than doubled (D) Less than doubled

Tardigrade Admin · Accepted Answer

Let E1 and E2 be the KE of photoelectrons for incident light of frequency v and 2v respectively. Then hv=E1+φ and h2v=E2+φ0 So, 2(E1 + (φ)0)=E2+(φ)0 or E2=2E1+φ0 It means the KE of photoelectron becomes more than double

Q. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is

Double the earlier value

Unchanged

More than doubled

Less than doubled

Let $E_{1}$ and $E_{2}$ be the KE of photoelectrons for incident light of frequency $v$ and $2 v$ respectively.
Then $h v = E_{1} + ϕ$ and $h 2 v = E_{2} + ϕ_{0}$
So, $2 (E_{1} + (ϕ)_{0}) = E_{2} + (ϕ)_{0}$ or $E_{2} = 2 E_{1} + ϕ_{0}$
It means the KE of photoelectron becomes more than double

Q. The frequency of the incident light falling on a photosensitive metal plate is doubled, the kinetic energy of the emitted photoelectron is

Double the earlier value

Unchanged

More than doubled

Less than doubled

Let E1​ and E2​ be the KE of photoelectrons for incident light of frequency v and 2v respectively. Then hv=E1​+ϕ and h2v=E2​+ϕ0​ So, 2(E1​+(ϕ)0​)=E2​+(ϕ)0​ or E2​=2E1​+ϕ0​ It means the KE of photoelectron becomes more than double

Let $E_{1}$ and $E_{2}$ be the KE of photoelectrons for incident light of frequency $v$ and $2 v$ respectively.
Then $h v = E_{1} + ϕ$ and $h 2 v = E_{2} + ϕ_{0}$
So, $2 (E_{1} + (ϕ)_{0}) = E_{2} + (ϕ)_{0}$ or $E_{2} = 2 E_{1} + ϕ_{0}$
It means the KE of photoelectron becomes more than double