Question

The ratio of energy required to remove an electron from first three bohrs orbit of hydrogen is

Answer 1

energy =-13.6×z^2/n^2

so  

energy of 1 orbital is: -13.6×1/1=-13.6

energy of 2 orbital is =-13.6×1/4

energy of 3 orbital is =-13.6×1/9

so its ratio will be

36:9:4

hope it helps

Answer 2

Answer:

The ratio of energy required to remove an electron from first three bohrs orbit of hydrogen is 1:4:9.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}eV$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Energy of the first orbit in hydrogen atom.

$E_1=-13.6\times \frac{1^2}{1^2} eV$

$E_2=-13.6\times \frac{1^2}{(2 )^2} eV=-3.4 eV$

$E_3=-13.6\times \frac{1^2}{(3)^2} eV=-1.51 eV$

$E_{\infty }=-13.6\times \frac{1^2}{(\infty )^2} eV=0 eV$

Energy required to remove an electron from first orbit of the hydrogen atom :

$E_1=E_{\infty }-E_1=0-(-13.6 eV)=13.6 eV$

Energy required to remove an electron from second orbit of the hydrogen atom :

$E_2=E_{\infty }-E_2=0-(-3.4 eV)=3.4 eV$

Energy required to remove an electron from third orbit of the hydrogen atom :

$E_3=E_{\infty }-E_3=0-(-1.51 eV)=1.51 eV$

The ratio of energy required to remove an electron from first three Bohr orbit of hydrogen :

$E_1:E_2:E_3$

$13.6 eV:4.4 eV:1.51 eV=1:4:9$