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Question

Physics Question on de broglie hypothesis

A hot filament liberates an electron with zero initial velocity. The anode potential is 1200 V1200\,V. The speed of the electron when it strikes the anode is

A

1.5×105 ms−11.5 \times 10^5 \,ms^{-1}

B

2.5×106 ms−12.5 \times 10^6 \,ms^{-1}

C

2.1×107 ms−12.1 \times 10^7 \,ms^{-1}

D

2.5×108 ms−12.5 \times 10^8 \,ms^{-1}

Answer

2.1×107 ms−12.1 \times 10^7 \,ms^{-1}

Explanation

Solution

12mv2=Vq\frac{1}{2} m v^{2}=V q v=2Vqmv=\sqrt{\frac{2 V q}{m}} =2.1×107ms−1=2.1 \times 10^{7} ms ^{-1}