Atoms And Nuclei Question 432

Question: In hydrogen atom, if the difference in the energy of the electron in n=2 and n=3 orbits is E, the ionization energy of hydrogen atom is

Options:

A) 13.2 E

B) 7.2 E

C) 5.6 E

D) 3.2 E

Show Answer

Answer:

Correct Answer: B

Solution:

(b) Energy =0.0258 amu (K = constant) n1 = 2 and n2 = 3, so

E=K[122132]=K[536]

For removing an electron n1 = 1 to n2= Energy E1=K[1]=365E=7.2 E

Ionization energy = 7.2 E



NCERT Chapter Video Solution

Dual Pane