According to Bohr’s theory, the electronic energy of H-atom in Bohr’s orbit is given by

Question

According to Bohr’s theory, the electronic energy of H-atom in Bohr’s orbit is given by (A) En=-(2.18 × 10-19 × Z/2n2) J (B) En=-(2.179 × 10-18 × Z2/n2) J (C) En=-(21.79 × 10-18 × Z/2n2) J (D) En=-(21.8 × 10-21 × Z2/n2) J

Tardigrade Admin · Accepted Answer

According to Bohr's theory, the energy of electron in

n^{th}

Bohr's orbit of

H

-atom or hydrogen-like atoms/ions is given by:

E_{n} = - R_{H} \frac{Z ^{2}}{n ^{2}}

where

R_{H} = 2.18 \times 1 0^{- 18} J

and

Z =

atomic number,

n = n^{th}

orbit
Thus,

E_{n} = - 2.18 \times 1 0^{- 18} \times \frac{Z ^{2}}{n ^{2}} J

Q. According to Bohr’s theory, the electronic energy of $H$ -atom in Bohr’s orbit is given by

$E_{n} = - \frac{2.18 \times 1 0 ^{- 19} \times Z}{2 n ^{2}} J$

$E_{n} = - \frac{2.179 \times 1 0 ^{- 18} \times Z ^{2}}{n ^{2}} J$

$E_{n} = - \frac{21.79 \times 1 0 ^{- 18} \times Z}{2 n ^{2}} J$

$E_{n} = - \frac{21.8 \times 1 0 ^{- 21} \times Z ^{2}}{n ^{2}} J$

Q. According to Bohr’s theory, the electronic energy of H-atom in Bohr’s orbit is given by

En​=−2n22.18×10−19×Z​J

En​=−n22.179×10−18×Z2​J

En​=−2n221.79×10−18×Z​J

En​=−n221.8×10−21×Z2​J

According to Bohr's theory, the energy of electron in nth Bohr's orbit of H-atom or hydrogen-like atoms/ions is given by: En​=−RH​n2Z2​ where RH​=2.18×10−18J and Z= atomic number, n=nth orbit Thus, En​=−2.18×10−18×n2Z2​J

Q. According to Bohr’s theory, the electronic energy of $H$ -atom in Bohr’s orbit is given by

$E_{n} = - \frac{2.18 \times 1 0 ^{- 19} \times Z}{2 n ^{2}} J$

$E_{n} = - \frac{2.179 \times 1 0 ^{- 18} \times Z ^{2}}{n ^{2}} J$

$E_{n} = - \frac{21.79 \times 1 0 ^{- 18} \times Z}{2 n ^{2}} J$

$E_{n} = - \frac{21.8 \times 1 0 ^{- 21} \times Z ^{2}}{n ^{2}} J$