For a photosensitive surface,threshold wave-length is `lambda_0`.Does photoemission occur,if the wavelength `(lambda)` off the incident radiation is
more than `lambda_0,`

For a given photosensitive material, how does photoelectric current change on decreasing the wavelength of incident radiation?

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

Find the value of lambda the equation is 5x+y+lambda=0 where x=1 and y=5

For constructive interfrence to take place between two monochromatic light waves of wavelength lambda ,the path difference should be

The figure shows the stopping potential for the phototelectron versus (1/lambda) graph for two metals A and B, lambda being the wavelength of incident light How is the value of Planck's constant determined from the graph?

The figure shows the stopping potential for the phototelectron versus (1/lambda) graph for two metals A and B, lambda being the wavelength of incident light If the distance between the light source and the surface of metal A is increased, how will the stopping potentail for the electrons emitted from it be affected? justify your answer.

The equation 2 sin^(3) theta+(2 lambda-3) sin^(2) theta-(3 lambda+2) sin theta-2 lambda=0