Question

The value of Planck's constant is $6.63\times {10}^{-34}Js$. The velocity of light is $3.0\times {10}^8$ ms${}^{-1}$. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of $8\times {10}^{15}{s}^{-1}$:

• A. $4\times {10}^1$
• B. $4\times {10}^4$
• C. $3\times {10}^3$
• D. $2\times {10}^2$

Answer 1

The correct option is A $4\times {10}^1$

Solution:-

As we know that,

$\nu =\frac{c}{\lambda }$

$\Rightarrow \lambda =\frac{c}{\nu }$

where,

$\lambda$ = wavelength of a quantum of light = ?

$\nu$ = frequency of a quantum of light = $8\times {10}^{15}{s}^{-1}$

$c$ = speed of light = $3\times {10}^{8}m/s$

$\therefore \lambda =\frac{3\times {10}^8}{8\times {10}^{15}}=0.375\times {10}^{-7}m=3.75\times {10}^{1}nm(\because 1m={10}^{9}nm)$

Hence, $4\times {10}^1$ will be the closest value to the wavelength (in nm) of a quantum of light with frequency $8\times {10}^{15}{s}^{-1}$.