Find the energy of each of the photons which

(i) correspond to light of frequency 3×1015 Hz.

(ii) have a wavelength of 0.50 Å.

Asked by Pragya Singh | 1 year ago |  76

1 Answer

Solution :-

(i) The energy of a photon (E) can be calculated by using the following expression:

E=\(h\nu\)

Where, ‘h’ denotes Planck’s constant, which is equal to 

\(6.626\times 10^{-34}\, Js\nu(frequency \;of \;the\; light) \)

\( = 3\times 10^{15}Hz\)

Substituting these values in the expression for the energy of a photon, E:

\(E=(6.626\times 10^{-34})(3\times 10^{15})\\ \\ E=1.988\times 10^{-18}\, J\)

(ii)

The energy of a photon whose wavelength is \(\lambda\) is:

E=\(hc\nu\)

Where,

h (Planck’s constant) 

=\(6.626\times 10^{-34}Js\)

c (speed of light) 

\(3\times 10^{8}\, m/s\)

Substituting these values in the equation for ‘E’:

\(E=\dfrac{(6.626\times 10^{-34})(3\times 10^{8})}{0.50\times 10^{-10}}\)

\( =3.976\times 10^{-15}J\\ \\ \)

\( ∴ E=3.98\times 10^{-15}J\)

Answered by Pragya Singh | 1 year ago

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