Ratio of frequency of revolution of electron in the second excited state of He - and second state of hydrogen isA- 32-27B- 27-32C- 27-2D- 1-54

The correct option is A Frequency v=V/2π r; V∝Z/n and r∝n^2/Z V=Z× Z/n× n ⇒ v ∝z^2/n^3 So v_He_2^⊕/v_H_2^⊕=2^2/3^2/1^2/2^2=8× 4/27=32/27 Second excited ⇒ n=3 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Positive Rays and Neutrons

Ratio of freq...

Question

Ratio of frequency of revolution of electron in the second excited state of $H e^{\oplus}$ and second state of hydrogen is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Step 1:
Frequency $v = \frac{V}{2 π r}; V \propto \frac{Z}{n} a n d r \propto \frac{n^{2}}{Z}$
$V = \frac{Z \times Z}{n \times n}$
Step 2:
$\Rightarrow v \propto \frac{z^{2}}{n^{3}}$
So $\frac{v_{H e_{2}^{\oplus}}}{v_{H_{2}^{\oplus}}} = \frac{\frac{2^{2}}{3^{2}}}{\frac{1^{2}}{2^{2}}} = \frac{8 \times 4}{27} = \frac{32}{27}$

Hence, option (A) is correct.
Second excited $\Rightarrow$ n=3

Suggest Corrections

Similar questions

Q. Ratio of frequency of revolution of electron in second excited state of

H e^{+}

and first excited state of

H

\frac{2^{x}}{3^{y}}

. Then the value of

x + y

is:

Q. The time period of revolution of

e^{-}

in ground state orbit of hydrogen atom is

T_{λ}

, then frequency of

e^{-}

in second excited state of

L i^{+ +}

ion is

Q. The time period of revolution of

e^{-}

in ground state orbit of hydrogen atom is

T_{λ}

, then frequency of

e^{-}

in second excited state of

L i^{+ +}

ion is

Q. An electron in H-atom jumps from second excited state to first excited state and then from first excited to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively. Then

Q. What is the ratio of time periods of an electron revolving in the second orbit of a hydrogen atom to another electron revolving in the third orbit of

H e^{+}

Join BYJU'S Learning Program

Ratio of frequency of revolution of electron in the second excited state of He⊕ and second state of hydrogen is

Step 1: Frequency v=V2πr; V∝Zn and r∝n2Z V=Z×Zn×nStep 2: ⇒v∝z2n3 So vHe⊕2vH⊕2=22321222=8×427=3227 Hence, option (A) is correct. Second excited ⇒ n=3

Ratio of frequency of revolution of electron in the second excited state of $H e^{\oplus}$ and second state of hydrogen is