Question

A gas absorbs a photon 355 nm and emits at two wavelenghts if one of the emmision is at 680 nm the other is at ??

Answer 1

where E1 is Energy of first emitted photon emitted and E2 is Energy of second emitted photon. Energy E and wavelength λ of a photon are related by the ... ⇒1λ2=680−355355×680

Answer 2

Explanation:

From Law of Conservation of energy, energy of absorbed photon must be equal to combined energy of two emitted photons.

E

T

=

E

1

+

E

2

.....(1)

where

E

1

is Energy of first emitted photon emitted and

E

2

is Energy of second emitted photon.

Energy

E

and wavelength

λ

of a photon are related by the equation

E

=

h

c

λ

.....(2)

where

h

is Planck's constant,

c

is velocity of light.

Inserting values from (2) in (1) we get

h

c

λ

T

=

h

c

λ

1

+

h

c

λ

2

⇒

1

λ

T

=

1

λ

1

+

1

λ

2

......(3)

Substituting given values in (3) we get

1

355

=

1

680

+

1

λ

2

⇒

1

λ

2

=

1

355

−

1

680

⇒

1

λ

2

=

680

−

355

355

×

680

⇒

λ

2

=

742.77

n

m