Question

A nucleus $X$ is in the higher energy state where its energy is $450keV$ . If it comes to ground state (assuming ground state energy to be zero), what is the wavelength of the photon emitted?

Answer 1

Given that :-

Energy $=E=450keV$

Ground state energy $=0$

$\therefore$ Energy of photon released $=E-0=450-0$

$\left(\Delta E\right)$

$\Delta E=450KeV=450\times {10}^{3}eV=450\times {10}^3\times 1.6\times {10}^{-19}J$

$\Delta E=450\times 1.6\times {10}^{-16}J$

$\Delta E=\frac{hC}{\lambda }$

$\lambda =$ wavelength $=\frac{hc}{\Delta E}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{450\times 1.6\times {10}^{-16}}$

$\lambda =2.76\times {10}^{-12}m=2.76pm$

Wavelength of photon emitted $=\lambda =2.76pm$