Calculate the radius of Bohr's first orbit for hydrogen atom and the energy of electron in this orbit:

-2.178

\times

^{- 18}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

21.78

\times

^{- 18}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.170

\times

^{- 18}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

21.88

\times

^{- 18}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A -2.178 $\times$ 10 $^{- 18}$ J
As we know,
r $_{n}$ = $\frac{n^{2} h^{2}}{4 π^{2} m e^{2} K}$ r $_{1}$
= $\frac{(1)^{2} (6.626 \times 10^{- 34})^{2}}{4}$ $\times (3.14)^{2} (9.109 \times 10^{- 31}) (1.6 \times 10^{- 9})^{2} (9 \times 10^{9})$ = 0.529 $\times$ 10 $^{- 10}$ = $0.529 A^{0}$
E = - $\frac{K z e^{2}}{2 r}$
= - $\frac{(9 \times 10^{9}) (1) (1.6 \times 10^{- 19})^{2}}{2 \times 0.529 \times 10^{- 10}}$
= $- 2.178 \times$ 10 $^{- 18}$ J

Suggest Corrections

Similar questions

The energy associated with the first orbit in the hydrogen atom is –2.18 × $10^{- 18}$ J $a t o m^{- 1}$ . What is the energy associated with the fifth orbit?

Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Q. The energy of an electron in the first Bohr's orbit of a hydrogen atom is

2.18 \times 10^{- 18} J

. Its energy in the second orbit will be:

Q. The energies of an electron in the first orbit of

H e^{+}

and in the third orbit of

L i^{2 +}

in joule are respectively:

Q. The ionization energy of the ground state of hydrogen atom is

2.18 \times 10^{- 18}

J. The energy of an electron in its second orbit would be:

Q. If the value of kinetic energy of an electron in

1 s t

orbit of a hydrogen atom is

2.18 \times 10^{- 18} J

, then what will be the energy of the electron in

5 t h

orbit?

Join BYJU'S Learning Program

Calculate the radius of Bohr's first orbit for hydrogen atom and the energy of electron in this orbit:

The correct option is A -2.178 × 10−18 JAs we know,rn = n2h24π2me2Kr1 = (1)2(6.626×10−34)24×(3.14)2(9.109×10−31)(1.6×10−9)2(9×109)= 0.529 × 10−10 = 0.529 A0E = -Kze22r= - (9×109)(1)(1.6×10−19)22×0.529×10−10= −2.178× 10−18 J