The Rydberg constant $R$ for hydrogen is

R = - (\frac{1}{4 π \in_{0}}) \frac{2 π^{2} m e^{2}}{c h^{2}}

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

R = (\frac{1}{4 π \in_{0}}) \frac{2 π^{2} m e^{4}}{c h^{2}}

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

R = {(\frac{1}{4 π \in_{0}})}^{2} \frac{2 π^{2} m e^{4}}{c^{2} h^{2}}

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

R = {(\frac{1}{4 π \in_{0}})}^{2} \frac{2 π^{2} m e^{4}}{c h^{3}}

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

Open in App

Solution

The correct option is D $R = {(\frac{1}{4 π \in_{0}})}^{2} \frac{2 π^{2} m e^{4}}{c h^{3}}$
$R = {[\frac{1}{4 π ϵ_{0}}]}^{2} . \frac{2 π^{2} m e^{4}}{c h^{3}}$

Because $R = \frac{2 π^{2} k^{2} e^{4} m}{c h^{3}} = {(\frac{1}{4 π \in_{0}})}^{2} \frac{2 π^{2} m e^{4}}{c h^{3}}$

Suggest Corrections

Similar questions

d B^{\to} = \frac{μ o}{4 π}

\frac{i d l^{\to} \times r^{\to}}{| r |^{3}}

sin θ + \sqrt{3} cos θ = 6 x - x^{2} - 11

0 \leq θ \leq 4 π, x \in R

V = \frac{4}{3} π r^{3}

\frac{d r}{d t} = 0.01

(ϕ_{r})

ϕ_{r} = \frac{q}{4 π ϵ_{0} r} . . . (r ⩾ R_{0})

ϕ_{r} = \frac{q}{4 π ϵ_{0} R_{0}} . . . (r \leq R_{0})

x sin x = 1,

4 π

4 π,

Join BYJU'S Learning Program