The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged a μ with mass 207 times that of an electron, is am equal to

Question

The radius of the hydrogen atom in its ground state is a0. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged a μ with mass 207 times that of an electron, is am equal to (A) 20 a0 (B) (a0/207) (C) (a0/√207) (D) a0√207

Tardigrade Admin · Accepted Answer

a0 = (h2ε0/π me2) aμ = (h2ε 0/π (207 m)e2) Dividing (ii) by (i), we get, (aμ /a0 ) = (1/207) or aμ = (a0 /207)

Q. The radius of the hydrogen atom in its ground state is $a_{0}$ . The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged $a_{μ}$ with mass 207 times that of an electron, is am equal to

$20 a_{0}$

$\frac{a _{0}}{207}$

$\frac{a _{0}}{207}$

$a_{0} 207$

$a_{0} = \frac{h ^{2} ε _{0}}{πm e ^{2}}$
$a_{μ} = \frac{h ^{2} ε _{0}}{π ( 207 m ) e ^{2}}$
Dividing $(ii)$ by $(i)$ , we get, $\frac{a _{μ}}{a _{0}} = \frac{1}{207}$ or $a_{μ} = \frac{a _{0}}{207}$

Q. The radius of the hydrogen atom in its ground state is a0​. The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged aμ​ with mass 207 times that of an electron, is am equal to

20a0​

207a0​​

207​a0​​

a0​207​

a0​=πme2h2ε0​​ aμ​=π(207m)e2h2ε0​​ Dividing (ii) by (i), we get, a0​aμ​​=2071​ or aμ​=207a0​​

Q. The radius of the hydrogen atom in its ground state is $a_{0}$ . The radius of a muonic hydrogen atom in which the electron is replaced by an identically charged $a_{μ}$ with mass 207 times that of an electron, is am equal to

$20 a_{0}$

$\frac{a _{0}}{207}$

$\frac{a _{0}}{207}$

$a_{0} 207$

$a_{0} = \frac{h ^{2} ε _{0}}{πm e ^{2}}$
$a_{μ} = \frac{h ^{2} ε _{0}}{π ( 207 m ) e ^{2}}$
Dividing $(ii)$ by $(i)$ , we get, $\frac{a _{μ}}{a _{0}} = \frac{1}{207}$ or $a_{μ} = \frac{a _{0}}{207}$