Radiation of wavelength 546 nm falls on a photo cathode and electrons with maximum kinetic energy of 0.18 eV are emitted. When radiation of wavelength 185 nm falls on the same surface, a (negative) stopping potential of 4.6 V has to be applied to the collector cathode to reduce the photoelectric current to zero. Then, the ratio `(h)/(e)` is

Options

A. `6.6xx10^(-15)JsC^(-1)`
B. `4.14xx10^(-15)JsC^(-1)`
C. `6.6xx10^(-34)JsC^(-1)`
D. `4.12xx10^(-34)JsC^(-1)`

Answer

B

Explanation

`h(v_1-v_2)=e(V_1-V_2)`
`(h)/(e)=(1)/(c)((V_1-V_2))/(((1)/(lamda_1)-(1)/(lamda_2)))=(10^(-9))/(3xx10^(8))((4.6-0.08)/((1)/(185)-(1)/(546)))`
`=(4.42xx185xx546xx10^(-17))/(361xx31)`
`=4.12xx10^(-15)JsC^(-1)`