Q27 / 120 · VITEEE 2017 · Chemistry
The values of Planck's constant is $6.63 \times 10^{-34} \mathrm{Js}$. The velocity of light is $3.0 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of $8 \times 10^{15}$ $\mathrm{s}^{-1} ?$
- A$5 \times 10^{-18}$
- Correct answerB$4 \times 10^{1}$
- C$3 \times 10^{7}$
- D$2 \times 10^{-25}$
Solution
$\mathrm{E}=\mathrm{h} \mathrm{v}=\frac{\mathrm{ch}}{\lambda} ;$ and $\mathrm{v}=\frac{\mathrm{c}}{\lambda}$
$8 \times 10^{15}=\frac{3.0 \times 10^{8}}{\lambda}$
$\begin{aligned}
\therefore \lambda &=\frac{3.0 \times 10^{8}}{8 \times 10^{15}}=0.37 \times 10^{-7} \\
&=37.5 \times 10^{-9} \mathrm{~m}=4 \times 10^{1} \mathrm{~nm}
\end{aligned}$
Hence, the correct answer is option: 'B'.