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?

A proton and alpha -particles are aceelerated through the same potential difference. The ratio of their de-Brogile wavelength will be:

Mention the significance of Davisson and Germer's experiment. An alpha-particl e and a proton are accelerated from rest through the same potential difference V. find the ratio of de-Broglie wavelength associated with them.

A proton, a neutron, an electron and alpha -particle have the same energy. Then their de-Broglie wavelengths compare is

An alpha partaicle is accelerated through a potential difference of 10^4 V . The gain in K.E. of the alpha particle is .

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

Be^(3+) and a proton are accelerated by the same potential, their de - Broglie wavelengths have the ratio ( assume mass of proton = mass of neutron ) :

de Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is :