In a photoelectric effect experiment, the kinetic energy of an emitted electron is `1.986 xx 10^(-19) J` , when a radiation of frequency `1.0 xx 10^15 s^(-1)` hits the metal. What is the threshold frequency of the metal (in `s^(-1)` )?(Planck's constant ` = 6.62 xx 10^(-34)` Js)

Options

A. `7.0 xx 10^14`
B. `5.8886 xx 10^14`
C. `7.0 xx 10^(-15)`
D. `7.0 xx 10^15`

Answer

A

Explanation

Given , kinetic energy of an emitted electron ` = 1.986 xx 10^(-19) J`
Frequency of radiation ` = 1 xx 10^15 s^(-1)`
Threshold frequency , `v_0 = ? `
From equation ,
`hv - hv_0 = KE`
` h(v - v_0) = KE`
`6.62 xx 10^(-34) J -s (1 xx 10^15 s^(-1) - v_0) = 19.986 xx 10^(-19) J`
` 1 xx 10^15 s^(-1) - v_0 = (1.986 xx 10^(-19) J)/(6.62 xx 10^(-34) J - s)`
`v_0 = 1 xx 10^15 s^(-1) -(1.986 xx 10^(-19) J)/(6.62 xx 10^(-34) J-s)`
`v_0 = 1 xx 10^15 - 0.3 xx 10^15 s^(-1)`
` v_0 = 7.0 xx 10^14 s^(-1)`