The acceleration of electron in the first orbit of a hydrogen atom is (where the symbols have their usual meanings)

Question

The acceleration of electron in the first orbit of a hydrogen atom is (where the symbols have their usual meanings) (A) (4 π 2 m/h3) (B) (h2/4 π 2 m r) (C) (h2/4 π 2 m2 r3) (D) (m2 h2/4 π 2 r3 )

Tardigrade Admin · Accepted Answer

From mvr=(n h/2 π ),v=(n h/2 π m r) For the first Bohr orbit, n=1 Acceleration, a=(v2/r)=(n2 h2/4 (π )2 m2 r2 (. r .))=(h2/4 (π )2 m2 r3)

Q. The acceleration of electron in the first orbit of a hydrogen atom is (where the symbols have their usual meanings)

$\frac{4 π ^{2} m}{h ^{3}}$

$\frac{h ^{2}}{4 π ^{2} m r}$

$\frac{h ^{2}}{4 π ^{2} m ^{2} r ^{3}}$

$\frac{m ^{2} h ^{2}}{4 π ^{2} r ^{3}}$

From $m v r = \frac{nh}{2 π}, v = \frac{nh}{2 πm r}$
For the first Bohr orbit, $n = 1$
Acceleration, $a = \frac{v ^{2}}{r} = \frac{n ^{2} h ^{2}}{4 ( π ) ^{2} m ^{2} r ^{2} ( r )} = \frac{h ^{2}}{4 ( π ) ^{2} m ^{2} r ^{3}}$

Q. The acceleration of electron in the first orbit of a hydrogen atom is (where the symbols have their usual meanings)

h34π2m​

4π2mrh2​

4π2m2r3h2​

4π2r3m2h2​

From mvr=2πnh​,v=2πmrnh​ For the first Bohr orbit, n=1 Acceleration, a=rv2​=4(π)2m2r2(r)n2h2​=4(π)2m2r3h2​

$\frac{4 π ^{2} m}{h ^{3}}$

$\frac{h ^{2}}{4 π ^{2} m r}$

$\frac{h ^{2}}{4 π ^{2} m ^{2} r ^{3}}$

$\frac{m ^{2} h ^{2}}{4 π ^{2} r ^{3}}$

From $m v r = \frac{nh}{2 π}, v = \frac{nh}{2 πm r}$
For the first Bohr orbit, $n = 1$
Acceleration, $a = \frac{v ^{2}}{r} = \frac{n ^{2} h ^{2}}{4 ( π ) ^{2} m ^{2} r ^{2} ( r )} = \frac{h ^{2}}{4 ( π ) ^{2} m ^{2} r ^{3}}$