# ENERGY AND WAVELENGTH OF PHOTON IIT JEE - NEET

## A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at :

(AIEEE 2010)

a) 325 nm

b) 743 nm

c) 518 nm

d) 1035 nm

Logic:

Since energy is conserved, the energy, Eabs of absorbed photon is equal to total of energies, (E1+ E2) of two photons emitted.

i.e. Eabs = E1+ E2

Where:

E1 = Energy of first photon emitted

E2 = Energy of second photon emitted

Energy, E and wavelength, λ of a photon are related as follows:

E = hc/λ

Where:

h = Planck's constant

c = velocity of light (a constant)

Therefore:

1/λabs = 1/λ1 + 1/λ2

Solution:

Upon substituting the known values in above equation:

1/355 = 1/680 + 1/λ2

By solving above equation:

λ2 = 742.76 nm

Conclusion:

Correct option is "b".

## Related questions

### 1) The energy of a photon is 3 x 10-12 erg. What is the wavelength in nm?

#### (Eamcet - 2006-E)

a) 662

b) 1324

c) 66.2

d) 6.62

Solution:

We know that:

E = hc/λ

or

λ= hc/E

Therefore:

wavelength λ = hc/E = 6.626 x 10-27 erg s x 3.0 x 1010 cm s-1 / 3 x 10-12 erg = 6.626 x 10-5 cm

Now:

1 cm = 107 nm

Therefore the wavelength in nm = 6.626 x 10-5 x 107 nm = 6.626 x 102 nm = 662.

#### (IIT JEE 1986)

a) 1/4

b) 4

c) 1/2

d) 2

Solution:

From

E = hc/λ

The ratio of energies of two radiations is given by:

E1/E2 = λ21 = 4000/2000 = 2

Note: Energy is inversely proportional to wavelength.

### 2) Which one of the following frequencies of radiation (in Hz) has a wavelength of 600 nm?

#### (EAMCET 2011-M)

a) 5.0 x 1014

b) 2.0 x 1013

c) 5.0 x 1016

d) 2.0 x 1014

Solution:

c = νλ

where

ν = frequency

λ  = wavelength

c = velocity of light

Therefore:

ν = c / λ = 3.0 x 108 m s-1 / 600 x 10-9 m =  5.0 x 1014 s-1 = 5.0 x 1014 Hz

## Homework

1) If the wavelength of an electromagnetic radiation is 2000  Å , what is the energy in ergs?

2) What is the wavenumber of electromagnetic radiation that has a wavelength 3 x 10-8 cm?

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