Question

Supposing the IP of hydrogen atom is 960eV. Find out the value of principal quantum number having energy equal to -60eV.

Answer 1

Answer:

En=E1/n^2 I.P=_E1

-60=-960/n^2

n^2 =96/6

n^2 =16

n=4

Answer 2

The value of principal quantum number having energy equal to -60 eV is 4

Explanation:

Ionization energy to ionize the electron from hydrogen atom will equal to the difference between energy of electron at infinity from the energy of electron at n = 1 energy level.

$I.E=E_{\infty }-E_1$

$960eV=0-E_1$

$E_1=-960eV$

$E_n=E_1\times \frac{Z^2}{n^2}eV$

where,

$E_n$ = energy of nth orbit = -60 eV  

n = number of orbit  = ?

Z = atomic number  = 1 (for hydrogen)

Putting in the values we get:

$-60=-960\times \frac{1^2}{n^2}eV$

$n^2=16$

$n=4$

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