Question

18) If the number of revolutions made by electron in 1.0 s in H-atom in its nth orbit is twice of
the number of revolutions made by electron in 1.0 s in the 2nd orbit of He ion, then 'n' is​

Answer 1

Answer:

n=number of revolution per second=circumferenceoforbitvelocityofe−

Vn=2.188×106×nzμn=0.529A∙×zn2

Putting z=1 and n=2

V2=2.188×106×21μ2=0.529×4×10−10=2.116×10−10mn=2×3.14×2.116×10−101.094×106=0.082×1016=8.2×1014revolution/sec

Answer 2

Answer:

The value of n is 2.

Explanation:

We know that the revolutions made per second are given as

$rps = \frac{v}{2\pi r}$, where v is the velocity of the electron.

The velocity of the electron is given as

$v=\frac{nh}{2\pi mr}$, where n is the Bohr orbit, h is Planck's constant, m is the mass, and r is the radius of the electron.

Therefore, the revolutions made per second are given as

$rps=\frac{nh}{4\pi^2 mr^2}$

It is given that the number of revolutions made by the electron in the hydrogen atom is twice that of the revolutions made in the helium atom.

$rps_1=2rps_2$

Substituting the formula for revolutions made per second.

$\frac{n_1h}{4\pi^2 mr_1^2} = 2 \frac{n_2h}{4\pi^2 mr_2^2}$

Since h and m are constants, we can cancel them. Therefore,

$\frac{n_1}{r_1^2} = 2 \frac{n_2}{r_2^2}$

$\frac{n_1}{(\frac{a_1n_1^2}{Z_1} )^{2} } =2 \frac{n_2}{(\frac{a_2n_2^2}{Z_2} )^{2} }$

$\frac{n_1Z_1^2}{a_1^2n_1^4} = 2\frac{n_2Z_2^2}{a_2^2n_2^4}$

$\frac{Z_1^2}{a_1^2n_1^3} =2 \frac{Z_2^2}{a_2^2n_2^3}$

Substituting $Z_1 = 1, Z_2=2, a_1=0.53*10^{-10}, a_2=1.4*10{-10}, n_2=2$ we get,

$\frac{1^2}{(0.53*10^{-10})^2n_1^3} =2 \frac{2^2}{(1.4*10^{-10})^22^3}$

$\frac{1^2}{(0.53*10^{-10})^2n_1^3} = \frac{2^3}{(1.4*10^{-10})^22^3}$

$\frac{1}{(0.53*10^{-10})^2n_1^3} = \frac{1}{(1.4*10^{-10})^2}$

$n_1^3 = \frac{(1.4*10^{-10})^2}{(0.53*10^{-10})^2}$

$n_1^3=\frac{1.96*10^{-20}}{0.2809*10^{-20}}$

$n_1^3 = 6.98$

$n_1^3 = 7$

$n_1 = 1.9$

$n_1 = 2$

Therefore, the electron is revolving in the second orbit of the hydrogen atom.

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