In a hydrogen atom, find the binding energy of an electron at n=4 state:

Question

In a hydrogen atom, find the binding energy of an electron at n=4 state: (A) 13.6eV (B) 3.4eV (C) 1.51eV (D) 0.85eV

Tardigrade Admin · Accepted Answer

answer Binding energy is nothing but the energy required to remove an electron from its orbit which is given by the equation En=(13 . 6/n2)eV ⇒ E4=(13 . 6/(4)2) ⇒ E4=0.85eV

Q. In a hydrogen atom, find the binding energy of an electron at $n = 4$ state:

$13.6 e V$

$3.4 e V$

$1.51 e V$

$0.85 e V$

answer
Binding energy is nothing but the energy required to remove an electron from its orbit which is given by the equation
$E_{n} = \frac{13.6}{n ^{2}} e V$
$\Rightarrow E_{4} = \frac{13.6}{( 4 ) ^{2}}$
$\Rightarrow E_{4} = 0.85 e V$

Q. In a hydrogen atom, find the binding energy of an electron at n=4 state:

13.6eV

3.4eV

1.51eV

0.85eV

answer Binding energy is nothing but the energy required to remove an electron from its orbit which is given by the equation En​=n213.6​eV ⇒E4​=(4)213.6​ ⇒E4​=0.85eV

Q. In a hydrogen atom, find the binding energy of an electron at $n = 4$ state:

$13.6 e V$

$3.4 e V$

$1.51 e V$

$0.85 e V$

answer
Binding energy is nothing but the energy required to remove an electron from its orbit which is given by the equation
$E_{n} = \frac{13.6}{n ^{2}} e V$
$\Rightarrow E_{4} = \frac{13.6}{( 4 ) ^{2}}$
$\Rightarrow E_{4} = 0.85 e V$