Question

electron in Hydrogen atom Jumped n= 3 to n = 1 find the energy and wavelength radiation.​

Answer 1

Answer:

energy released=12.09eV

wavelength=1/9.6536×$10^{8}$ m

Explanation:

E3⟶2​=E3​−E1

​E3​=−​$\frac{13.6}{9}$=−1.51eV

E1​=$\frac{13.6}{1}$=−13.6eV

Therefore, energy released is,  E2- E1= 13.6-1.51=12.09eV

1/ λ​=(1.097×$10^{7}$) (1-$\frac{1}{9}$)

1/ λ=(1.097×$10^{7}$)($\frac{8}{9}$)

1/ λ=(1.097×$10^{7}$) (0.88)

1/ λ=9.6536×$10^{8}$

λ=1/9.6536×$10^{8}$ m