Question

A light bulb $14\text{W}$ emits white light, which is a mixture of various colours in the spectrum. The energy is distributed among the seven colours i.e., red, orange, yellow, green, purple, indigo and blue. If the number of photons $n$ emitted from each colour is equal, then the frequency of green colour is (The frequency of all the colours of white light are in AP)

$\left(\frac{n}{\text{time}}\right)=5.5\times {10}^{18}{\text{s}}^{-1}$

• A. $5.5\times {10}^{14}\text{Hz}$
• B. $6.2\times {10}^{12}\text{Hz}$
• C. $5.2\times {10}^{16}\text{Hz}$
• D. $5.4\times {10}^{17}\text{Hz}$

Answer 1

The correct option is A $5.5\times {10}^{14}\text{Hz}$

Given: $P=14\text{W}$

As we know, $E=nh\nu$

$P\times t=nh\nu$

$n$ is same for all the light waves so

$14=\left(\frac{n}{t}\right)\left[h\right][{\nu }_1+{\nu }_2+{\nu }_3+{\nu }_4+{\nu }_5+{\nu }_6+{\nu }_7]$

Green colour radiation is the middle one in the $\text{AP}$, so,

$2{\nu }_4={\nu }_3+{\nu }_5={\nu }_2+{\nu }_6={\nu }_1+{\nu }_7$

$\therefore 14=5.5\times {10}^{18}[6.6\times {10}^{-34}]\left[7{\nu }_4\right]$

$\Rightarrow {\nu }_4=5.5\times {10}^{14}\text{Hz}$

Hence, option $\left(\text{A}\right)$ is correct.

Key Concept: The total energy of white light is the sum of $\left(h\nu \right)$'s of individual monochromatic light.