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PhysicsModern PhysicsAtomic structureMedium2 minPYQ_2023
PhysicsMediumnumerical

An atom absorbs a photon of wavelength500 nmand emits another photon of wavelength600 nm. The net energy absorbed by the atom in this process isn×10-4eV. The value of n is [Assume the atom to be stationary during the absorption and emission process] (Takeh=6.6×10-34J sandc=3×108 m s-1).

Answer:
4125.00
Solution:

The energy of a photon is given by E=hcλ.

It is given that λ1=500 nmλ2=600 nm.

The net energy absorbed is 

ΔE=hcλ1-hcλ2=hc10-9(1500-1600)

=6.6×10-34×3×108×100500×600×10-9

=6.6×330×10-19 J

1 eV=1.6×10-19 J

=6.6×330×1.6eV

=4125×10-4 eV

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure
2mℹ️ Source: PYQ_2023

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