An electron in the hydrogen atom jumps from excited state `n` to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function `2.75eV`. If the stopping potential of the photoelectron is `10V`, the value of `n` is

Options

A. 13.4 eV
B. 12.75eV
C. 5.1 eV
D. 2.31 eV

Answer

B

Explanation

`KE_(max) = 10 eV`
` phi = 2.75 eV `
total incident energy
` E= phi + KE _( max ) = 12.75 e V `
` therefore ` energy is released when electron jumps from the excited state n to the groud state .
` : . E_4 -E_1 = { -0.85 - ( 13.6 ) eV } `
`=12.75eV `