Question

The Planck’s constant is $6 \cdot 63 \times {10^{ - 34}}Js$. The velocity of light is $3 \cdot 0 \times {10^8}m{s^{ - 1}}$. Which value is closest to the wavelength in nanometre 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

Hint: We are going to use $\lambda = \dfrac{c}{u}$ where, $\lambda $ is wavelength, $c$ is velocity of light in vacuum, $u$ is frequency of light. A photon is a single quantum of light . quantum also told us that light consists of both light and matter which is of tiny particles which possess wave-like properties.

Complete answer:
In the question, given values are Planck’s constant $h = 6 \cdot 63 \times {10^{34}}Js$, velocity of light in vacuum is $c = 3 \cdot 0 \times {10^8}m{s^{ - 1}}$ and the frequency of a quantum of light is $u = 8 \times {10^{15}}{s^{ - 1}}$
All it takes is a simple formula to solve the question above, here the Planck’s constant is generally given to confuse or make the student think deep into the question. So therefore, Planck’s constant is not used anywhere in solving the equation. So, the equation goes like,
$\lambda = \dfrac{c}{u}$
So, all we have to find here is the wavelength in nanometre and compare it with the $4$ options associated with the question and find out the closest value to it. So, therefore we have to find the wavelength. The value of $c$ and $u$ is directly provided in question.
$c = 3 \cdot 0 \times {10^8}m{s^{ - 1}}$
$u = 8 \times {10^{15}}{s^{ - 1}}$
$\lambda = \dfrac{{3 \cdot 0 \times {{10}^8}m{s^{ - 1}}}}{{8 \times {{10}^{15}}{s^{ - 1}}}}$
Therefore, the value of wavelength becomes,
$\lambda = 37 \cdot 5 \times {10^{ - 9}}m$
As it is the ${10^{ - 9}}$ it becomes nanometre,
$\lambda = 37 \cdot 5nm$
Now that we have obtained the value in nanometre,
We need to now compare with options above,
Let us first expand the value
$4 \times {10^1} = 40$
$4 \times {10^4} = 40,000$
$3 \times {10^3} = 3,000$
$2 \times {10^2} = 200$
So, clearly in the above expansion it is clear that option (A) $40$ is closest to the value of wavelength $37 \cdot 5nm$ .

Hence, the correct option is (A) .

Note:
The emission of electrons from the surface of material following the adsorption of electromagnetic radiation . Electromagnetic radiations consist of radiation which is oscillating in the electric and magnetic field oriented perpendicular to each other moving in the space .