CBSE Class 12 · 2016 · Delhi · Modern PhysicsShort Answer

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(i) State Bohr's quantization condition for defining stationary orbits. How does the de-Broglie hypothesis explain the stationary orbits?
(ii) Find the relation between the three wavelengths

λ_{1}

λ_{2}

and

λ_{3}

from the energy level diagram shown.

Answer: $m v r = nh /2 π$ ( $2 π r = nλ$ ); $1/ λ_{3} = 1/ λ_{1} + 1/ λ_{2}$ .

Solution

(i) Bohr's condition: the electron can occupy only those orbits in which its angular momentum is an integral multiple of

\frac{h}{2 π}

, i.e.

m v r = \frac{nh}{2 π}

. Using de-Broglie's hypothesis (

λ = h / m v

), this is equivalent to

2 π r = nλ

: a stationary orbit is one whose circumference contains a whole number of de-Broglie wavelengths, forming a standing wave that does not radiate.
(ii) From the level diagram (levels

A

B

C

with

C > B > A

E_{C} - E_{B} = \frac{h c}{λ _{1}}

E_{B} - E_{A} = \frac{h c}{λ _{2}}

E_{C} - E_{A} = \frac{h c}{λ _{3}}

. Adding the first two and equating to the third:

\frac{h c}{λ _{1}} + \frac{h c}{λ _{2}} = \frac{h c}{λ _{3}} \Rightarrow \frac{1}{λ _{3}} = \frac{1}{λ _{1}} + \frac{1}{λ _{2}} .

\frac{1}{λ _{3}} = \frac{1}{λ _{1}} + \frac{1}{λ _{2}}

CBSE Class 12 Board · 2016 · Delhi · Subjective

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CBSE Class 12 · 2016 · Delhi · Modern PhysicsShort Answer

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