Question

18. An electron is moving in nth orbit of radius 2.3758A
in He+ ion. Energy required to remove electron from
this orbit of Het is
(1) 6.04eV
(2) 5.45eV
(4) 6.0253
(3) 54.4eV​

Answer 1

Explanation:

there may be some calculation error so the excat value is not coming but its coming close

Answer 2

Answer:(1)6.04ev

Explanation: Given that the the electron is moving in the nth orbit and the radius of that orbit is 2.3758A in He+ ion(Z=2) as per Bohr model and also that the energy needed to remove it is needed to be found out:-

So, we say that:-

The formula for the nth orbit of the radius is 0.529×n²/Z,

So here its give for the nth orbit the value is 2.3758A.

The value is 0.529×n²/2(For He+)=2.3758 So,

Both are in Angstroms.

So, n²=2.3758×2/0,529=9.

So, value of n is 3.

After getting the value of n we use the formula of nth state which is:-

En=-13.6×Z²/n² so here we know the values of Z and n just put it in the formula for the energy:-

It is:-

En=-13.6×4/9(Putting n and z as 3 and 2 and squaring them):-

So the energy is 6.04ev.

Or the ground state energy of:-

H atom is 13.6ev and He is 54.4 ev that is always constant so just divide by n² to 54.4ev.

So it is 54.4/9=6.04 Answer.

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