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

A) 13.2E

B)7.2E

C)5.6E

D)3.2E

1 Answer

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answered by (9.5k points)

we know the energy of electron in n th orbit of Hydrogen

En = -K/n^2ev, k = constant

So the difference in the energy of the electron in n=2 and n=3 orbits is

E = E3 – E= -K[1/3^2-1/2^2] = K*5/36

=>K = 7.2E

Now ionization energy of hydrogen atom is the energy required to promote electron from n=1  to n=

So This ionization energy = -K{1/∞^2-1/1^2]=K =7,2E

Hence option (B) is accepted

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