Find the ratio of energy for radiations having $6000 \ \mathring{A}$ and $4000 \ \mathring{A}$ wavelength.
$(h = 6.62 \times 10^{-34} \ J \ s, c = 3 \times 10^8 \ m \ s^{-1})$

The energy of one mole of photons of radiation of frequency $2 \times 10^{12} \ Hz$ in $J \ mol^{-1}$ is $.....$ (Nearest integer) (Given: $h = 6.626 \times 10^{-34} \ Js$,$N_A = 6.022 \times 10^{23} \ mol^{-1}$)

Electromagnetic radiation of wavelength $663 \ nm$ is just sufficient to ionise the atom of metal $A.$ The ionization energy of metal $A$ in $kJ \ mol^{-1}$ is ...... .
(Rounded-off to the nearest integer) $\left[ h = 6.63 \times 10^{-34} \ Js, c = 3.00 \times 10^{8} \ ms^{-1}, N_{A} = 6.02 \times 10^{23} \ mol^{-1} \right]$

If the energies of two photons are in the ratio of $3:2$,their wavelengths will be in the ratio of

In which type of radiation do the mass number and atomic number remain unchanged?

Which of the following is not an electromagnetic radiation?