When a certain metal was irradiated with light of frequency `40xx10^(16) s^(-1)`, the photoelectrons emitted had four times kinetic energy as the kinetic energy of photoelectrons emitted when the same metal was irradiated with light of frequency `2.0 xx 10^(16) s^(-1)`. the trhreshold frequency (`v_(0)`) of the metal in `s^(-1)` is

Options

A. `2xx10^(16)`
B. `4 xx 10^(16)`
C. `2.5 xx 10^(16)`
D. `1.33 xx 10^(16)`

Answer

D

Explanation

Given, `V_(1) = 4 xx 10^(16) sec^(-1)`
`v_(2) = 2xx 10^(16) sec^(-1)`
`KE_(1) = 4 KE_(2)`
From equation of photoelectric effect ,
` hv = hv_(0) + KE " "` …..(i)
`therefore " " hv_(1) = hv_(0) + 4 KE " " (v_(1) = 2v_(2) )`
`hv_(2) = hv_(0) + KE " " ` .... (ii)
From Eq. (i) and (ii) , we have
` hv_(2) = 3 KE " " [ because v_(1) = 2v_(2) ] `
On putting the above calculated value in eq (ii) ,
we get
Now, the maximum range of this transmission depends upon the height of transmitting antenna and is given by d = `sqrt(2h R_(e))`.
where, h = height of tower
and `R_(e)` = Radius of earth (R `gt gt` h )
So, the number of people received the transmission,
Population covered, (`P_(c)`) population density (n) `xx ` area of transmission range (A)
Population covered, `(P_(c )) = n xx A `
`P_(c ) = (1000)/(pi) xx pi d^(2)`
Area of transmission range,
A = ` pi d^(2) or P_(c ) = 1000 xx 2h R_(e ) " " [ because d^(2) = 2h R_(e)]`
or` P_(c) =1000 xx 2 xx 5 xx 10^(-3) xx 6400`
or ` P_(c ) = 64000`
So, the number of people received the transmission is 64000 .
`(2hv_(2))/(3) = hv_(0)`
` (2 xx h xx 2 xx 10^(16))/(3 ) = hv_(0)`
`v_(0) = 1.33 xx 10^(16)`