Chemistry, asked by kalyani7353, 4 months ago

4h
1) 20
2) 20
3) 27
Passage-3: A formula analogous to the Rydberg formula applies to the series of spectral lines
which arise from transitions from higher energy levels to the lower energy level of
hydrogen atom.
A muonic hydrogen ion is like a hydrogen atom in which the electron is replaced by a
heavier partical the muon. The mass of the muon is 207 times the mass of an electron while
charge the double that the electron (assume charge of proton is same)
7. Radius of first Bohr orbit of muonic hydrogen ion is
0.529

4) 0.529 x 828 A
8. Ionization energy of muonic hydrogen ion is
1) 13.6 x 207 eV 2) 13.6 x 414 eV 3) 13.6 x 828 eV
9. Distance between 1st and 3rd Bohr orbit of muonic hydrogen ion will be
4) 13.6 x 3312 eV
-X8 4°
x8A
-X54°
2) 414
X8A
4) 828
0.529
0.529

A
1) 207
2) 414
3) 828
0.529
0.529
0.529
0.529
1) 207
3) 207​

Answers

Answered by mohddaniyal794
0

Answer:

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11th

Chemistry

Structure of Atom

Shapes and Energies of Atomic Orbitals

A formula analogous to the ...

Chemistry

A formula analogous to the Rydberg formula applies to the series of spectral lines which arise from transitions from higher energy level to the lower energy level of hydrogen atom.

A muonic hydrogen atom is like a hydrogen atom in which the electron is replaced by a heavier particle, the 'muon'. The mass of the muon is about 207 times the mass of an electron, while the charge remains same as that of the electron. Rydberg formula for hydrogen atom is:

λ

1

=R

H

[

n

1

2

1

n

2

2

1

](R

H

=109678cm

−1

)

Energy of first Bohr orbit of muonic hydrogen atom is

Medium

Answer

Correct option is

B

−13.6×207eV

E∝m

m is the mass of the particle (electron or muon) and E is the energy of first orbit of the atom.

E

electronic

E

muonic

=

m

electron

m

muon

−13.6eV

E

muonic

=

1

207

r

muonic

=−13.6×207eV

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