If an em wave of wavelength `lambda` is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength `lambda_1`, prove that
`lambda=((2mc)/h)lambda_1^2`

Options

A. `sqrt((2mc)/(hlamda))`
B. `sqrt((hlamda)/(2mc))`
C. `sqrt((hc)/(2mlamda))`
D. `(2mc)/(h lamda)`

Answer

B

Explanation

As the work function is negligible,
`therefore K.E. " of photoelectron = Energy of the incident wave" `
`therefore 1/2 mv^2 = hv = (hc)/(lamda)`
`therefore mv^2 = (2hc)/(lamda)`
`therefore (m^2v^2)/(m) =(2hc)/(lamda) therefore mv = sqrt((2hcm)/(lamda))`
`therefore` de broglie wavelength ` lamda ' = (h)/(mv)`
`= (h)/(sqrt((2hcm)/(lamda)))`
`therefore lamda' = sqrt((h^2lamda)/(2hcm)) = sqrt((hlamda)/(2mc))`