Comment
Share
Q)

# The threshold frequency of the metal of the cathode in a photoelectric cell is $1 \times 10^{15}\: Hz$ . When a certain beam of light is incident on the cathode, it is found that a stopping potential '4.144 V' is required to reduce the current to zero. The frequency of the incident radiation is : $( h = 6.63 \times 10^{-34} J-s)$

$\begin {array} {1 1} (1)\;2.5 \times 10^{15}\: Hz & \quad (2)\;2 \times 10^{15}\: Hz \\ (3)\;4.144 \times 10^{15}\: Hz & \quad (4)\;3 \times 10^{16}\: Hz \end {array}$

Comment
A)
Ans : (2)
$2 \times 10^{15}\: Hz$

Comment
A)
Give full solution of above question how to get this answer

Comment
A)

(h=plank constant) f,frequency f0,threshold frequency

ev0=hf-hf0 (v0=stopping potten

tial) f=ev0/h+v0