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Exercises · 2.6

Q.Find energy of each of the photons which

(i) correspond to light of frequency 3×10153 \times 10^{15} Hz.
(ii) have wavelength of 0.50 Å.
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Photon energy is directly proportional to frequency and inversely proportional to wavelength. Using E=hνE = h\nu and E=hc/λE = hc/\lambda, the energies are 1.99×10−181.99 \times 10^{-18} J for the given frequency and 3.98×10−153.98 \times 10^{-15} J for the given wavelength.

The core idea here is beautifully simple: a photon is a quantum of light, and its energy is locked to its electromagnetic wave properties. You cannot talk about a photon's energy without its frequency or wavelength — they are two sides of the same coin. Planck's constant hh is the bridge that connects the wave picture to the particle picture.

For part (i), we have frequency directly, so the relation E=hνE = h\nu is the most natural path. For part (ii), we are given wavelength, so we use E=hc/λE = hc/\lambda. Both are equivalent because ν=c/λ\nu = c/\lambda for light in vacuum.

Let's work through each part step by step.

  1. Part (i): Using frequency directly The energy of a photon is given by Planck's relation:

E=hνE = h\nu

where h=6.626×10−34 J⋅sh = 6.626 \times 10^{-34} \ \text{J·s} (Planck's constant) and ν=3×1015 Hz\nu = 3 \times 10^{15} \ \text{Hz}.

Substituting:

E=(6.626×10−34)×(3×1015)E = (6.626 \times 10^{-34}) \times (3 \times 10^{15})

Multiply the numbers: 6.626×3=19.8786.626 \times 3 = 19.878, and the powers of ten: 10−34×1015=10−1910^{-34} \times 10^{15} = 10^{-19}.

So:

E=19.878×10−19 J=1.9878×10−18 JE = 19.878 \times 10^{-19} \ \text{J} = 1.9878 \times 10^{-18} \ \text{J}

Rounding to three significant figures (matching the given data), we get 1.99×10−181.99 \times 10^{-18} J.

Tip

A quick mental check: light in the visible range has frequencies around 101410^{14} Hz and photon energies around 10−1910^{-19} J. Here the frequency is 3×10153 \times 10^{15} Hz — ten times higher — so the energy should be about ten times larger, which it is.

  1. Part (ii): Using wavelength The wavelength is given as 0.500.50 Å. Recall that 1 A˚=10−10 m1 \ \text{Å} = 10^{-10} \ \text{m}, so:

λ=0.50×10−10 m=5.0×10−11 m\lambda = 0.50 \times 10^{-10} \ \text{m} = 5.0 \times 10^{-11} \ \text{m}

The energy-wavelength relation is:

E=hcλE = \frac{hc}{\lambda}

where c=3.0×108 m/sc = 3.0 \times 10^8 \ \text{m/s} (speed of light).

First compute hchc:

hc=(6.626×10−34)×(3.0×108)=1.9878×10−25 J⋅mhc = (6.626 \times 10^{-34}) \times (3.0 \times 10^8) = 1.9878 \times 10^{-25} \ \text{J·m}

Now divide by λ\lambda:

E=1.9878×10−255.0×10−11=1.98785.0×10−25+11=0.39756×10−14 JE = \frac{1.9878 \times 10^{-25}}{5.0 \times 10^{-11}} = \frac{1.9878}{5.0} \times 10^{-25+11} = 0.39756 \times 10^{-14} \ \text{J}

Which is:

E=3.9756×10−15 JE = 3.9756 \times 10^{-15} \ \text{J} …

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