If muonic hydrogen atom is an atom in which a negatively charged muon (μ) of mass about 207 me revolves around a proton, then first Bohr radius of this atom is (re = 0.53 × 10-10 m), if ground state energy of electron is -13.6 eV then what is the ground state energy of muonic hydrogen atom?

Question

If muonic hydrogen atom is an atom in which a negatively charged muon (μ) of mass about 207 me revolves around a proton, then first Bohr radius of this atom is (re = 0.53 × 10-10 m), if ground state energy of electron is -13.6 eV then what is the ground state energy of muonic hydrogen atom? (A) 1.8 keV (B) -2.8 keV (C) -3.8 keV (D) 4.8 keV

Tardigrade Admin · Accepted Answer

According to Bohrâs atomic model energy, E propto m ∴ (Eμ/Ee) = (mμ/me) Here, Ee = -13.6 eV ; mμ = 207 me ∴ Eμ =(mμ/me)Ee = (207me/me ) ×(-13.6)eV = -2815.2 eV = -2.8 keV

$1.8 k e V$

$- 2.8 k e V$

$- 3.8 k e V$

$4.8 k e V$

According to Bohr’s atomic model energy,
$E \propto m$
$∴ \frac{E _{μ}}{E _{e}} = \frac{m _{μ}}{m _{e}}$
Here, $E_{e} = - 13.6 e V;$
$m_{μ} = 207 m_{e}$
$∴ E_{μ} = \frac{m _{μ}}{m _{e}} E_{e}$
$= \frac{207 m _{e}}{m _{e}} \times (- 13.6) e V$
$= - 2815.2 e V$
$= - 2.8 k e V$

Q. If muonic hydrogen atom is an atom in which a negatively charged muon (μ) of mass about 207me​ revolves around a proton, then first Bohr radius of this atom is (re​=0.53×10−10m), if ground state energy of electron is −13.6eV then what is the ground state energy of muonic hydrogen atom?

1.8keV

−2.8keV

−3.8keV

4.8keV

According to Bohr’s atomic model energy, E∝m ∴Ee​Eμ​​=me​mμ​​ Here, Ee​=−13.6eV; mμ​=207me​ ∴Eμ​=me​mμ​​Ee​ =me​207me​​×(−13.6)eV =−2815.2eV =−2.8keV

$1.8 k e V$

$- 2.8 k e V$

$- 3.8 k e V$

$4.8 k e V$

According to Bohr’s atomic model energy,
$E \propto m$
$∴ \frac{E _{μ}}{E _{e}} = \frac{m _{μ}}{m _{e}}$
Here, $E_{e} = - 13.6 e V;$
$m_{μ} = 207 m_{e}$
$∴ E_{μ} = \frac{m _{μ}}{m _{e}} E_{e}$
$= \frac{207 m _{e}}{m _{e}} \times (- 13.6) e V$
$= - 2815.2 e V$
$= - 2.8 k e V$