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Question

Physics Question on Atoms

An HH -atom moving with speed vv makes a head-on collision with another HH -atom at rest. Both the atoms are in the ground state. The minimum value of velocity vv for which one of the atoms may excite is

A

6.25×104ms1\text{6.25}\times 10^{4}ms^{- 1}

B

8×104ms1\text{8}\times 10^{4}ms^{- 1}

C

7.25×104ms1\text{7.25}\times 10^{4}ms^{- 1}

D

13.6×104ms1\text{13.6}\times 10^{4}ms^{- 1}

Answer

6.25×104ms1\text{6.25}\times 10^{4}ms^{- 1}

Explanation

Solution

For vminv_{m i n} collision should be completely inelastic. According to energy conservation principle, 12mvmin2=12mv2+12mv2+ΔE\therefore \frac{1}{2}mv_{m i n}^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}mv^{2}+\Delta E ...(i)) According to momentum conservation principle. mvmin=mv+mvmv_{m i n}=mv+mv v=mmin2\therefore v=\frac{m_{m i n}}{2} ...(ii) After solving e (i) an d(ii), we get 12mvmin2=2ΔE\frac{1}{2}mv_{m i n}^{2}=2\Delta E vmin=(4ΔEm)\therefore v_{m i n}=\sqrt{\left(\frac{4 \Delta E}{m}\right)} Here, ΔE=\Delta E= minimum excitation energy =10.2eV=10.2 \, eV m=1.67×1027kgm=1.67\times 10^{- 27} \, kg vmin=(4×10.2×1.6×(10)191.67×(10)27)\therefore v_{m i n}=\sqrt{\left(\frac{4 \times 10.2 \times 1.6 \times \left(10\right)^{- 19}}{1.67 \times \left(10\right)^{- 27}}\right)} =6.25×104ms1=\text{6.25}\times 10^{4}ms^{- 1}