Number of de Broglie waves made by a Bohr electron in an orbit of maximum magnetic quantum number $+ 2$ is:

1

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3

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Solution

The correct option is C $3$
The maximum magnetic quantum number $= + 2$

$m$ values are from $- l$ to $l$

This means $l = 2$

But, $l = 0$ to $n - 1$

Therefore, $2 = n - 1$

or $n = 3$

No. of waves formed $=$ Principal quantum number of the orbit

$∴$ No of waves is $3$ .

Hence, the correct option is $C$

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Number of de Broglie waves made by a Bohr electron in an orbit of maximum magnetic quantum number +2 is:

The correct option is C 3The maximum magnetic quantum number =+2m values are from −l to lThis means l=2But, l=0 to n−1Therefore, 2=n−1or n=3No. of waves formed = Principal quantum number of the orbit∴ No of waves is 3.Hence, the correct option is C

Number of de Broglie waves made by a Bohr electron in an orbit of maximum magnetic quantum number $+ 2$ is: