  Question

According to Bohr's model of hydrogen atom, relation between principal quantum number n and radius of stable orbit is

A
r1n  B
rn  C
r1n2  D
rn2  Solution

The correct option is D $$r\propto n^2$$Let an electron (mass m), is revolving around nucleus of a hydrogen atom in an orbit of radius $$r$$ with velocity $$v$$. Electron gets centripetal force to revolve, from the electrostatic attraction force between electron and nucleus,          $$\dfrac{mv^{2}}{r}=\dfrac{e\times e}{4\pi\varepsilon_{0}r^{2}}$$   (for H- atom, $$Z=1$$ hence charge on nucleus is $$=e$$)          $$mv^{2}=\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$  ...................eq1According to Bohr's model, angular momentum of electron :          $$mvr=\dfrac{nh}{2\pi}$$  ..................eq2Now, squaring eq2 and dividing by eq1, we get          $$\dfrac{mv^{2}r^{2}}{mv^{2}}=\dfrac{n^{2}h^{2}4\pi\varepsilon_{0}r}{4\pi^{2}e^{2}}$$            $$r=\dfrac{n^{2}h^{2}\varepsilon_{0}}{\pi me^{2}}$$  which gives $$r\propto n^{2}$$Physics

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