The velocity of a proton is $\frac{c}{20}$. The de Broglie wavelength associated to it is: (Here $c$ is the speed of light.)

A hydrogen atom moving at a speed $v$ absorbs a photon of wavelength $122 \mathrm{nm}$ and stops. Find the value of $v$.

(Mass of hydrogen atom$=1.67\times {10}^{-27 }\mathrm{kg}$)

If momentum of a particle is increased by $25\%$, then its wavelength:

Two particles having masses $m_1$ and $m_2$ have identical charges. If they are accelerated through identical potential difference, then the ratio of their de Broglie wavelengths is $\frac{{\lambda }_1}{{\lambda }_2}$ then: