Question

Calculate the energy associated with photon of light having a wavelength $6000\mathring{A}$. $[h=6.624\times {10}^{-27}ergsec]$.

Answer 1

We know that, $E=hv=h\cdot \frac{c}{\lambda }$
$h=6.624\times {10}^{-27}erg-sec$; $c=3\times {10}^{10}cm/sec$;
$\lambda =6000\mathring{A}=6000\times {10}^{-8}cm$
So, $E=\frac{(6.624\times {10}^{-27})\times (3\times {10}^{10})}{6\times {10}^{-5}}=3.312\times {10}^{-12}erg$.