Question

The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation.

Solution

Energy of photon, E = hv

This implies,

E = h \frac{C}{λ}

where,

h = 6.62 \times 10^{- 34} js a n d, c = 3 \times 10^{8} m s^{- 1}

If wavelength

λ

is in metre and energy is in Joule, then we will divide E by

1.6 \times 10^{- 19}

to convert into eV (electron volt).

∴

E = \frac{hc}{λ \times 1.6 \times 10^{- 19}} eV

(1) For y-rays, wavelength ranges from 10^-10 m to less that 10 ^-14 m.

∴

Energy =

\frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{10^{- 10} \times 1.6 \times 10^{- 19}} eV

= 12.4 \times 10^{3} eV \approx 10^{4} eV

Thus for,

λ = 10^{- 10} m, energy = 10^{4} eV .

and

λ = 10^{- 14} m, energy = 10^{8} eV .

i.e., Energy of

γ - rays

ranges between

10^{4} to 10^{8} eV .

(2) For X-rays, wavelength ranges from 10^-8 m to 10^-13 m.

For

λ = 10^{- 8}

∴

Energy = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{10^{- 8} \times 1.6 \times 10^{- 19}} eV = 124 \approx 10^{2} eV

λ = 10^{- 13} m, energy = 10^{7} eV .

(3) For ultraviolet radiations, λ ranges from 4 × 10^-7 m to 6 × 10^-10 m.

Therefore,

For

λ = 4 \times 10^{- 7}

Energy = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{4 \times 10^{- 7} \times 1.6 \times 10^{- 19}} eV = 3.1 eV \approx 10^{10} eV

λ = 6 \times 10^{- 10} m

, Energy = 10³ eV

Energy of ultraviolet radiations vary between

10^{10} to 10^{3} eV .

(4) For visible radiations, wavelength ranges from 4 × 10^-7 m to 7 × 10^-7 m.

Therefore,

For

λ = 4 \times 10^{- 7},

energy

= 10^{10} eV

(same as above)

For,

λ = 7 \times 10^{- 7}

;

Energy = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{7 \times 10^{- 7} \times 1.6 \times 10^{- 19}} eV

= 1.77 eV \approx 10^{0} eV

(5) For infrared radiations, λ ranges from 7 × 10^-7 m to 7 × 10^-14 m.

Therefore,

For

λ = 7 \times 10^{- 7}, E nergy = 10^{0} eV (as proved above)

For

λ = 7 \times 10^{- 4}

, energy is

\frac{1}{1000} times

.
i.e., of the order of

10^{- 3} eV .

(6) For micro waves, λ ranges from 1 mm to 0.3 m.
For

λ = 1 mm or 10^{- 3},

Energy is equal to

E = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{10^{- 3} \times 1.6 \times 10^{- 19}} eV

= 1.24 \times 10^{- 3} eV \approx 10^{- 3} eV .

For

λ = 0.3 m, Energy = 4.1 \times 10^{- 6} eV \approx 10^{- 6} eV .

(7) For radio waves, λ ranges from 1 m to few km.
For

λ = 1 m

,
Energy is equal to

E = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{10^{0} \times 1.6 \times 10^{- 19}} eV

= 1.24 \times 10^{- 6} eV \approx 10^{- 6} eV .

Energy for λ of the order of few km ≈ 10^-6 ev.

Energy of a photon that a source produces indicates, the spacing of relevant energy levels of the source.

For Daily Free Study Material Join wiredfaculty

Current Electricity

Some More Questions From Current Electricity Chapter

What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current A

Voltage V

Current A

Voltage V

0.2

0.4

0.6

0.8

1.0

2.0

3.94

7.87

11.8

15.7

19.7

39.4

3.0

4.0

5.0

6.0

7.0

8.0

59.2

78.8

98.6

118.5

138.2

158.0

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction