Question

What is the number of photons of light with a wavelength of 4000pm that provide 1J of energy?

Answer 1

The answer to this question includes calculation of energy of photons with the frequency given and the formula is as follows$E=\dfrac{hc}{\lambda }$ and then finding total number of photons present.

Complete step by step solution:
We have studied about photochemistry and also about the energy of photons in our lower chemistry classes.
Now let us study in detail about the energy of photons and its number.
- Energy of a photon is defined as the energy carried by a single photon and this energy possessed is directly proportional to the photon’s electromagnetic frequency and inversely proportional to the wavelength. Therefore, higher the frequency of a photon, higher will be the energy. Now from the given data we have wavelength given as $\lambda =4000pm=4000\times {{10}^{-12}}=4\times {{10}^{-9}}m$
Also energy of a single photon can be given by the formula which relates energy with the frequency and wavelength which is given as,
$E=\dfrac{hc}{\lambda }$
where, E is energy of the photon
h is the Planck’s constant
c is the velocity of light
and $\lambda$is the wavelength of light
Now, by substituting the values of these constants and also the values from the given data we get,
$E=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4\times {{10}^{-9}}}$
By solving we have,
$\Rightarrow E=4.969\times {{10}^{-17}}J$
This gives the energy of the single photon.
Now let us calculate the number of photons of light that provide 1J of energy and is given by,
No. of photons =$\dfrac{1J}{4.969\times {{10}^{-17}}J}$
$\Rightarrow n=2.012\times {{10}^{17}}$

Therefore, the correct answer is $2.012\times {{10}^{17}}$photons provide 1J of energy.

Note: The number of electrons emitted per second can also be calculated if frequency is given as frequency is inversely proportional to wavelength and the same formula can be written as $E=hc\nu$where$\nu$ is the frequency.