Question

# In Bohr's model of hydrogen atom, the electron circulates round the nucleus in a path of radius $$5\times 10^{-11}$$m at a frequency of $$6.8\times 10^{15}$$ revolutions per second. The value of magnetic induction at the centre of the orbit is :

A
12.27T
B
10.8T
C
13.2T
D
13.6T

Solution

## The correct option is A $$13.6T$$ $$\left | B \right |$$  =$$\dfrac{\mu _{0} i}{2r}$$I=$$\dfrac{\mathrm{d} q}{\mathrm{d} t}$$= $$qf$$= $$1.6\times 10^{-19}\times 6.8\times 10^{15}$$=$$10.88\times 10^{-4}$$$$\therefore \left | B \right | = \dfrac{4\pi\times 10^{-7}\times 10.88\times 10^{-4}}{2 \times 5 \times 10^{-11}}$$=$$13.6T$$Physics

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