(C) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v = 2.18 \times 10^6 \times \frac{Z}{n} \ ms^{

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(C) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v = 2.18 \times 10^6 \times \frac{Z}{n} \ ms^{-1}$.For a hydrogen atom,the atomic number $Z = 1$.For the third orbit,the principal quantum number $n = 3$.Substituting these values into the formula:$v = 2.18 \times 10^6 \times \frac{1}{3} \ ms^{-1}$.$v = 0.7266 \times 10^6 \ ms^{-1}$.$v = 7.26 \times 10^5 \ ms^{-1}$.Therefore,option $(C)$ is the correct answer.

Class 11 Chemistry Structure of Atom Atomic models and Planck's quantum theory

Medium

The speed of the electron (in $ms^{-1}$) in the third orbit of a hydrogen atom is approximately (mass of electron $= 9.1 \times 10^{-31} \ kg$).

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A
$3.6 \times 10^5$
B
$2.18 \times 10^6$
C
$7.26 \times 10^5$
D
$2.18 \times 10^5$

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The speed of the electron (in $ms^{-1}$) in the third orbit of a hydrogen atom is approximately (mass of electron $= 9.1 \times 10^{-31} \ kg$).

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What will be the frequency of the radiation emitted when an electron in a hydrogen atom jumps from $n = 4$ to $n = 1$? $(h = 6.625 \times 10^{-34} \,J \cdot s)$

The limiting line in the Balmer series will have a frequency of (Rydberg constant,$R_{\infty} = 3.29 \times 10^{15} \ s^{-1}$)

Assertion : Angular momentum of an electron in any orbit is given by angular momentum $= \frac{nh}{2\pi}$,where $n$ is the principal quantum number.Reason : The principal quantum number,$n$,can have any integral value.

In Bohr's atomic model,the radius $(r)$ and energy $(E)$ of an orbit $(n)$ are related as:

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Assertion : Angular momentum of an electron in any orbit is given by angular momentum $= \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
Reason : The principal quantum number,$n$,can have any integral value.