Question

Wavelengths of different radiations are given below:
$\lambda \left(A\right)=400\text{nm}\lambda \left(B\right)=400\mu \text{m}$
$\lambda \left(C\right)=40\text{nm}\lambda \left(D\right)=40\stackrel{\circ }{A}$
Arrange these radiations in the increasing order of their energies:

• A. $A<B<C<D$
• B. $C<B<D<A$
• C. $B<A<C<D$
• D. $C<D<A<B$

Answer 1

The correct option is C $B<A<C<D$

Given,
$\lambda \left(A\right)=400\text{nm}=4\times {10}^{-7}\text{m}$
$\lambda \left(B\right)=400\mu \text{m}=4\times {10}^{-4}\text{m}$
$\lambda \left(C\right)=40\text{nm}=4\times {10}^{-8}\text{m}$
$\lambda \left(D\right)=40\stackrel{\circ }{A}=4\times {10}^{-9}\text{m}$
We know, $E=hv=\frac{hc}{\lambda }$
Thus, the energy of radiation is inversely proportional to its wavelength.
$E\propto \frac{1}{\lambda }$
So, larger the value of wavelength, lower will be its energy.
Hence the order will be: $B<A<C<D$