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

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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