An electron, e1 is moving in the fifth stationary state, and another electron e2 is moving in the fourth stationary state. The radius of orbit of electron, e1 is five times the radius of orbit of electron, e1 calculate the ratio of velocity of electron e1(v1) to the velocity of electron e2 (v2).

From the expression of Bohr’s theory, we know that mev1r1=n1h2π&mev2r2=n2h2πmev1r1mev2r2=n1n2h2π×2πh Given, r1=5r2,n1=5,n2=4me×v1×5r2me×v2×r2=54⇒v1v2=54×5=14=1:4

An electron, e1 is moving in the fifth stationary state, and another electron e2 is moving in the fourth stationary state. The radius of orbit of electron, e1 is five times the radius of orbit of electron, e1 calculate the ratio of velocity of electron e1(v1) to the velocity of electron e2 (v2).

From the expression of Bohr’s theory, we know that mev1r1=n1h2π&mev2r2=n2h2πmev1r1mev2r2=n1n2h2π×2πh Given, r1=5r2,n1=5,n2=4me×v1×5r2me×v2×r2=54⇒v1v2=54×5=14=1:4

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An electron, e₁ is moving in the fifth stationary state, and another electron e₂ is moving in the fourth stationary state. The radius of orbit of electron, e₁ is five times the radius of orbit of electron, e₁ calculate the ratio of velocity of electron e₁(v₁) to the velocity of electron e₂ (v₂).

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1:4

5:1

1:5

4:1

answer is D.

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Detailed Solution

From the expression of Bohr’s theory, we know that

$\begin{array}{l} m_{e} v_{1} r_{1} = n_{1} \frac{h}{2 π} \\ & m_{e} v_{2} r_{2} = n_{2} \frac{h}{2 π} \\ \frac{m_{e} v_{1} r_{1}}{m_{e} v_{2} r_{2}} = \frac{n_{1}}{n_{2}} \frac{h}{2 π} \times \frac{2 π}{h} \\ Given, r_{1} = 5 r_{2}, n_{1} = 5, n_{2} = 4 \\ \frac{m_{e} \times v_{1} \times 5 r_{2}}{m_{e} \times v_{2} \times r_{2}} = \frac{5}{4} \\ \Rightarrow \frac{v_{1}}{v_{2}} = \frac{5}{4 \times 5} = \frac{1}{4} = 1 : 4 \end{array}$