The energy required to excite an electron in hydrogen atom to its first excited state is

Question

The energy required to excite an electron in hydrogen atom to its first excited state is (A) 8.5 eV (B) 10.2 eV (C) 12.7 eV (D) 13.6 eV

Tardigrade Admin · Accepted Answer

Energy, En = -(13.6/n2) eV In ground state energy E1 = -(13.6/12) = - 13.6eV In first excited state energy E2 = -(13.6/22) = - 3.4 eV Then the required energy = E2 - E1 = - 3.4 - (-13.6) = 10.2 eV

Q. The energy required to excite an electron in hydrogen atom to its first excited state is

$8.5 e V$

$10.2 e V$

$12.7 e V$

$13.6 e V$

Energy, $E_{n} = - \frac{13.6}{n ^{2}} e V$
In ground state energy $E_{1} = - \frac{13.6}{1 ^{2}}$
$= - 13.6 e V$
In first excited state energy $E_{2} = - \frac{13.6}{2 ^{2}}$
$= - 3.4 e V$
Then the required energy
$= E_{2} - E_{1}$
$= - 3.4 - (- 13.6)$
$= 10.2 e V$

Q. The energy required to excite an electron in hydrogen atom to its first excited state is

8.5eV

10.2eV

12.7eV

13.6eV

Energy, En​=−n213.6​eV In ground state energy E1​=−1213.6​ =−13.6eV In first excited state energy E2​=−2213.6​ =−3.4eV Then the required energy =E2​−E1​ =−3.4−(−13.6) =10.2eV

$8.5 e V$

$10.2 e V$

$12.7 e V$

$13.6 e V$

Energy, $E_{n} = - \frac{13.6}{n ^{2}} e V$
In ground state energy $E_{1} = - \frac{13.6}{1 ^{2}}$
$= - 13.6 e V$
In first excited state energy $E_{2} = - \frac{13.6}{2 ^{2}}$
$= - 3.4 e V$
Then the required energy
$= E_{2} - E_{1}$
$= - 3.4 - (- 13.6)$
$= 10.2 e V$