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Consider 3rd orbit of He^+ (Helium), usi...

Consider 3rd orbit of `He^+` (Helium), using non-relativistic approach , the speed of electron in this orbit will be [given `k=9xx10^(9)` constant, Z=2 and h(Planck's constant) =6.6`xx10^(-34)J*s`]

A

`2.92xx10^6 m//s`

B

`1.46xx10^6 m//s`

C

`0.73xx10^6m//s`

D

`3.0xx10^8m//s`

Text Solution

Verified by Experts

The correct Answer is:
B

According to Bohr's quantum condition,
`mv_nr_n=n(h)/(2pi) or, v_n=(hn)/(2pimr_n)`
`"Again",r_n prop (n^2)/(mZ) or,mr_n prop(n^2)/Z,`
`"Hence" ,v_n prop n(Z)/(n^2) i.e., v_n prop Z/n`
For ground state of hydrogen , Z=1 and n=1,
`therefore v=c/(137)`
`therefore "For third orbit of "He^+(Z=2)`,
`v=c/(137)*2/3=(3xx10^(8))/(137)xx2/3=1.46xx10^6 m//s`
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