Result Analysis Summary
Question 12 of 32 | Bohr's Model of Atom | Q4 - Q20
The radius of the first permitted Bohr orbit for the electron in a hydrogen atom equals 0.5 A
˚
and its ground state energy equals
−13.6 eV. If the electron in the hydrogen atom is replaced by a muon (μ )[charge same as electron and mass 207 m ], the first Bohr
−
e
radius and ground state energy will be:
( m represents the mass of an electron)
e
1. m, eV
−13
0.53 × 10 − 3.6
2. 25.6 × 10
−13
m, − 2.8 eV
3. 2.56 × 10
−13
m, − 2.8 keV
4. 2.56 × 10
−13
m, − 13.6 eV
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Explanation:
Hint: r
2 2
n h
=
2 2
4π mkze
Step: Find the first Bohr radius and ground state energy.
The reduced mass is given by;
mM
μ =
m+M
The Bohr radius is given by;
2 2
n h
r =
2 2
4π mkze
The reduced mass is written as;
me Mnucleus
μ =
me +Mnucleus
207me ×1836me
⇒ μ =
207me +1836me
⇒ μ = 186me
The radius of the first orbit is written as;
me
r1 = × 0.51
186me
−13
⇒ r1 = 2.56 × 10 m
and energy of the first Bohr is written as;
μ
E1 = E
m
186me
⇒ E1 = × (−13.6)
me
⇒ E1 = −2.8 keV
Hence, option (3) is the correct answer.
1x 00:00
(1) 12% (2) 14%
(3) 51% (4) 25%