Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6} imes10^5$ m s⁻¹. The radius of the orbit is $x imes10^{-9}$ m. The value of x is: [Take the mass of electron to be $9 imes10^{-31}$ kg, charge of electron = $-1.6 imes10^{-19}$ C and $\frac{1}{4\pi\varepsilon_0}=9 imes10^9$ N m² C⁻²]

Question

Consider that an electron is revolving in an excited state of Hydrogen atom with velocity $\sqrt{25.6}	imes10^5$ m s⁻¹. The radius of the orbit is $x	imes10^{-9}$ m. The value of x is: [Take the mass of electron to be $9	imes10^{-31}$ kg, charge of electron = $-1.6	imes10^{-19}$ C and $\frac{1}{4\pi\varepsilon_0}=9	imes10^9$ N m² C⁻²]

Dr. Abubacker · Accepted Answer

[Solution from source] Coulomb force gives the centripetal force: $r=ke^2/(mv^2)$. $v^2=2.56	imes10^{11}$, $ke^2=2.304	imes10^{-28}$, $mv^2=2.304	imes10^{-19}$, so $r=1.0	imes10^{-9}$ m and x = 1. Answer (1).

Explanation

Reviewed by Dr. Abubacker

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