The number of photons of light with the wavelength of 4000 pm that provides 1 J of energy is

  • Option 1)

    2\times10^{16}

  • Option 2)

    1\times10^{18}

  • Option 3)

    3\times10^{15}

  • Option 4)

    3\times10^{16}

 

Answers (1)

As learnt in

The energy (E) of a quantum of radiation -

E=hv

Where h is plank’s constant and \nu is frequency

-

 

 Thus, E= nhv

\Rightarrow 1 J = n\times 6.62\times 10^{-34}\times \frac{3\times 10^{8}}{4000\times 10^{-12}}

\Rightarrow n = 2\times 10^{16}


Option 1)

2\times10^{16}

This solution is correct.

Option 2)

1\times10^{18}

This solution is incorrect.

Option 3)

3\times10^{15}

This solution is incorrect.

Option 4)

3\times10^{16}

This solution is incorrect.

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