A particle momentum asscociated with de-Brogile wavelength `lambda` is `lambda=frac(h)(mv)`. (True/ false)

Calculate the de-Brogile wavelength of an electron.

The wavelength of a photon and its de-Brogile wavelength are_______

A proton and an alpha-particle are accelerated using the same potential difference. How are the de-Broglie wavelengths lambda_p and lambda_alpha related to each other?

If photon and an electron have same de-Brogile wavelength then.

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

According to de-Brogile hypothesis matter has particle nature. (True/ false)

State and explain de Brogile hypothesis for duality of matter and calculate the de-Brogile wavelength of an electron (i.e. a material particle).

Photon is not a material particle. (True/ false)

Plot a graph showing variation of de-Broglie wavelength (lambda) associated with a charged particle of mass m, versus 1/sqrtV , where particle is accelerated. How does this graph give us the information regarding the magnitude of the charge of the particle?