In a hydrogen atom- the electron is making 6-6- 1015 revs-1 around the nucleus in an orbit of radius 0-528 oA- The magnetic moment A-m2 will be-

The correct option is D 1× 10^ 23 Current, I=6.6× 10^15× 1.6× 10^ 19 =10.5× 10^ 4A .Area A=π R^2=3.142× (0.528)^2× 10^ 20m^2 so, Magnetic m ...

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Standard XII

Physics

Characteristics of Bohr's Atomic Model

In a hydrogen...

Question

In a hydrogen atom, the electron is making $6.6 \times 10^{15} r e v s^{- 1}$ around the nucleus in an orbit of radius $0.528 o A$ . The magnetic moment $(A - m^{2})$ will be.

1 \times 10^{- 15}

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1 \times 10^{- 10}

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1 \times 10^{- 23}

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1 \times 10^{- 27}

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Solution

The correct option is D $1 \times 10^{- 23}$
Current, $I = 6.6 \times 10^{15} \times 1.6 \times 10^{- 19}$
$= 10.5 \times 10^{- 4} A$ .
Area $A = π R^{2} = 3.142 \times (0.528)^{2} \times 10^{- 20} m^{2}$
so, Magnetic moment
$M = I A = 10.5 \times 10^{- 4} \times 3.142 \times (0.528)^{2} \times 10^{- 2}$
$= 10 \times 10^{- 24} = 10^{- 23}$ units

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In a hydrogen atom, the electron is making 6.6×1015revs−1 around the nucleus in an orbit of radius 0.528oA. The magnetic moment (A−m2) will be.

The correct option is D 1×10−23Current, I=6.6×1015×1.6×10−19=10.5×10−4A.Area A=πR2=3.142×(0.528)2×10−20m2so, Magnetic momentM=IA=10.5×10−4×3.142×(0.528)2×10−2=10×10−24=10−23 units

In a hydrogen atom, the electron is making $6.6 \times 10^{15} r e v s^{- 1}$ around the nucleus in an orbit of radius $0.528 o A$ . The magnetic moment $(A - m^{2})$ will be.