A monochromatic light source of frequency villuminates a metallic surface and ejects photoelectron. The electron with maximum energy are just able to ionize the hydrogen atom in the ground state. When the whole experiment is repeated with an incident radiation of frequency 5/6 v, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of the wavelength `1215Å`. Find the work function of the metal :

Options

A. 6.875 eV
B. 8.756 eV
C. 7.568 eV
D. 5.687 eV

Answer

A

Explanation

In first case
`hv=w+(1)/(2)mv_(1)^(2)=w+` Ionization energy
Now, ionisation potential =13.6V
`:.hv=w+1.6xx10^(-19)xx13.6`
`=w21.+76xx10^(-19) " "...(1)`
In second case
`h.(5)/(6)=w+(1)/(2) mv_(2)^(2)=w+(hc)/(lambda)`
`=w+(6.6xx10^(-34)xx3xx10^(8))/(1215xx10^(-10))`
`=w+16.3xx10^(-19) " "....(2)`
Dividing (1) by (2),
`(6)/(5)=(w+21.76xx10^(19))/(w+16.3xx10^(-19))`
or `w=11xx10^(-19)J=6.875eV`