Chapter 5, Problem 49A

Interpretation Introduction

Interpretation:

Energy of a photon of red light that has a frequency of $4.48\times {10}^{14}Hz$ needs to be calculated.

Concept introduction:

The energy of each quantum is directly proportional to the frequency ( $\nu$ ) of the radiation, i.e.

$E\propto \nu$ or $E=hv=\frac{hc}{\lambda }$

Where, $h=$ Planck's constant $=6.62\times {10}^{-27}erg.\sec .$ or $6.62\times {10}^{-34}Joules\sec .$

Answer to Problem 49A

Energy of photon $=29.68\times {10}^{-20}$ Joule

Explanation of Solution

Given information:

The frequency is $=4.48\times {10}^{14}Hz$

The relation between energy, Planck’s constant and frequency is as follows:

$E=h\nu$

Here,

$E=$ energy of photon

$h=$ Planck’s constant $\left(6.626\times {10}^{-34}Js\right)$

$\nu =$ frequency

Now, putting the values,

$E=6.626\times {10}^{-34}Js\times 4.48\times {10}^{14}Hz$

$E=6.626\times 4.48\times {10}^{-34+14}$

$E=2.968\times {10}^{-21}Joule$.

Conclusion

Energy of a photon of red light that has a frequency of $4.48\times {10}^{14}Hz$ is $2.968\times {10}^{-21}Joule$