The threshold wavelength of a photosensitive metal is 275nm.calculate the maximum velocity of the photoelectron ejected when light of 180nm is incident on it. 

 

​​​​​​​How to solve this?? 

Asked by Hemanthrakshitha95 | 3rd Mar, 2020, 02:55: PM

Expert Answer:

K E subscript m a x end subscript equals E minus ϕ
K E subscript m a x end subscript equals fraction numerator 1240 n m minus e V over denominator 180 n m end fraction minus fraction numerator 1240 n m minus e V over denominator 275 n m end fraction
K E subscript m a x end subscript equals 6.88 e V minus 4.55 e V
K E subscript m a x end subscript equals 2.33 e V
1 half cross times 9.31 cross times 10 to the power of negative 31 end exponent cross times v squared equals 2.33 cross times 1.6 cross times 10 to the power of negative 19 end exponent
v equals 9.05 cross times 10 to the power of 5 m divided by s

Answered by Utkarsh Lokhande | 3rd Mar, 2020, 03:55: PM