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Question

For He+the electron is in orbit with energy equal to 3.4eV. The azimuthal quantum number for that orbit is 2 and the magnetic quantum number is 0. Then which of the following is/are correct?


A

The subshell is 4d.

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B

The number of angular nodes in it is 2.

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C

The numbers of radial nodes in it is 3

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D

The nuclear charge experienced in n=4 is 2e less than that in n=1, where e is the electric charge.

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Solution

The correct option is B

The number of angular nodes in it is 2.


Explanation for the correct option:

Step-1

The energy of electron in Bohr's orbit:

E=Eo×Z2n2

where, Z is the atomic no. of Hydrogen like species

n is the principal quantum number

l is azimuthal quantum number

Step 2: Putting all the variables in above equation:

3.4=13.6×4/n2

After solving, we get:

n=4l=2

Step-3:

Therefore, from the above calculation it is clear that:

(A) The subshell is 4d

(B) Number of angular nodes =l

Number of angular nodes is 2

Explanation for the incorrect option:

(C) Number of radial nodes n-l-1

Number of radial nodes =4-2-1=1

Therefore, option (C) is incorrect.

(D) Nuclear charge would be the same hence, option (D) is incorrect

Therefore, the correct answer options are (A), (B).


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