Question

jee mains related questions​

Answer 1

Answer:

4th option

= -30. 6 eV

Explanation:

Formulae being used :-

1. Energy of an electron in nth orbit E = $-13.6 \times\frac{Z^{2} }{n^{2} }$
2. Radius of nth orbit = $0.529 \times \frac{n^{2} }{Z}$

Z being Atomic Number and n being the orbit number

Answering :-

Radius of nth orbit = 1.587 A

For Li²+ , Z = 3

Step I

Finding n using formula 2

r = $0.529 \times \frac{n^{2} }{Z}$

1.587 =  $\frac{n^{2} }{3}$

(Cross multiplying)

n² = 1.587 × 3

n = $\sqrt{1.587 \: \times \: 3}$

$= \: \sqrt{4.761}\\= \: 2.181$~ 2

n = 2

Step II

Finding Energy using formula 1

E = $-13.6\: \times\frac{Z^2}{n^2}$

$=-13.6 \: \times \frac{3^2}{2^2} \\=\frac{-13.6 \: \times 9}{4} \\= -30. 6$

E = -30. 6 eV

                         

Hope this helps!