When the light of frequency `2 v_(0)` (where `v_(0)` is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is `v_(1)`. When the frequency of the incident radiation is increased to `5 v_(0)`, the maximum velocity of electrons emitted from the same plate is `v_(2)`. the ratio of `v_(1)` to `v_(2)` is

Options

A. `4:1`
B. `1:4`
C. `1:2`
D. `2:1`

Answer

C

Explanation

According to the Einstein's photoelectric equation,
`K_(max) = 1/2mv_(max)^(2)= hv-ph_(0)=hv-hv_(0)`
where, `K_(max)` is the maximum kinetic energy of photoelectrons having maximum velocity `v_(max)`.
when incident frequency of light, v=`2v_(0)`
Substituting the value of v inn Eq (i), we get ltbr. `1/2mv_(1)^(2)=h(2v_(0))-hv_(0) = 2hv_(0) - hv_(0)=hv_(0).............(ii)
On dividing Eq. (ii) by Eq. (iii), we get
`(1/2mv_(1)^(2))/(1/2mv_(2)^(2))=(hv_(0))/(4hv_(0))rArr v_(1)^(2)/v_(2)^(2)=1/4 or v_(1)/v_(2)=1/2`
`therefore v_(1) : v_(2) = 1:2`