When $\beta$-particles are sent through a tin metal foil,most of them go straight through the foil as

$A$ quantum will have more energy if :

What is the work function of the metal if the light of wavelength $4000 \, \mathring{A}$ generates photoelectrons of velocity $6 \times 10^5 \, ms^{-1}$ from it? ............. $eV$ (Mass of electron $= 9 \times 10^{-31} \, kg$; Velocity of light $= 3 \times 10^8 \, ms^{-1}$; Planck's constant $= 6.626 \times 10^{-34} \, Js$; Charge of electron $= 1.6 \times 10^{-19} \, C \, eV^{-1}$)

An electron in an atom jumps in such a way that its kinetic energy changes from $x$ to $\frac{x}{4}$. The change in potential energy will be

The ratio of the energy of the electrons in the ground state of hydrogen to the electrons in the first excited state of $Be^{3+}$ is:

For a $H$ atom,the wavelength for the highest energy transition is $91.2 \ nm$. Calculate the corresponding wavelength for $He^{+}$ in $nm$.