If `lambda_(0)` is the threshold wavelength for photoelectric emission. `lambda` wavelength of light falling on the surface on the surface of metal, and `m` mass of electron. Then de Broglie wavelength of emitted electron is :-

Options

A. `[(h(lambdalambda_(0)))/(2mc(lambda_(0)-lambda))]overset(1)(2)`
B. `[(h(lambda_(0)-lambda))/(2mclambdalambda_(0))]overset(1)(2)`
C. `[(h(lambda-lambda_(0)))/(2mclambdalambda_(0))]overset(1)(2)`
D. `[(hlambdalambda_(0))/(2mc)]overset(1)(2)`

Answer

A

Explanation

`m_(n)` = mass of neutron , `m_(p)` = mass of proton
`(m_(n))/(2)" "2m_(p)`
atmoic mass `implies (m_(n)+m_(p))" "[m_(n)~=m_(p)] implies (8+6)=14m_(p)`
atomic mass `implies (4+12)=16m_(p)`
`%` increase `=(16-14)/(14)xx100=14.28%`