Question 12 Marks
Wavelengths of different radiations are given below:
$
\lambda=(A) 300 nm
$
$\lambda(B)=300 \mu m$
$\lambda( C )=3 nm$
$\lambda$ (D) $=30 \stackrel{\circ}{A}$
Arrange these radiations in the increasing order of their energies.
$
\lambda=(A) 300 nm
$
$\lambda(B)=300 \mu m$
$\lambda( C )=3 nm$
$\lambda$ (D) $=30 \stackrel{\circ}{A}$
Arrange these radiations in the increasing order of their energies.
Answer
View full question & answer→$
\begin{aligned}
& \lambda(A)=300 nm=300 \times 10^{-9} m \\
& \lambda(B)=300 \mu m=300 \times 10^{-6} m
\end{aligned}
$
$
\begin{aligned}
& \lambda(C)=3 nm=3 \times 10^{-9} m \\
& \lambda(D)=30 \stackrel{\circ}{A}=3 \times 10^{-9} m
\end{aligned}
$
We know that, $E \propto \frac{1}{\lambda}$.
Since, increasing order of wavelength is given as, $300 \mu m<300 nm<3 nm<30 \stackrel{o}{A}$
Therefore, $\lambda(B)<(A)<\lambda(C)=\lambda(D)$
\begin{aligned}
& \lambda(A)=300 nm=300 \times 10^{-9} m \\
& \lambda(B)=300 \mu m=300 \times 10^{-6} m
\end{aligned}
$
$
\begin{aligned}
& \lambda(C)=3 nm=3 \times 10^{-9} m \\
& \lambda(D)=30 \stackrel{\circ}{A}=3 \times 10^{-9} m
\end{aligned}
$
We know that, $E \propto \frac{1}{\lambda}$.
Since, increasing order of wavelength is given as, $300 \mu m<300 nm<3 nm<30 \stackrel{o}{A}$
Therefore, $\lambda(B)<(A)<\lambda(C)=\lambda(D)$
