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Question: The frequency of revolution of electron in n$^{th}$ Bohr orbit is $v_n$. The graph between log n and...

The frequency of revolution of electron in nth^{th} Bohr orbit is vnv_n. The graph between log n and log (vn/v1v_n/v_1) may be

A

(A)

B

(B)

C

(C)

D

(D)

Answer

C

Explanation

Solution

The frequency of revolution of an electron in the nth^{th} Bohr orbit, denoted as vnv_n, is given by the formula:

vn=velocity of electron in n-th orbitcircumference of n-th orbit=vn2πrnv_n = \frac{\text{velocity of electron in n-th orbit}}{\text{circumference of n-th orbit}} = \frac{v_n}{2 \pi r_n}

We know the expressions for the velocity (vnv_n) and radius (rnr_n) of an electron in the nth^{th} Bohr orbit for a hydrogen-like atom (atomic number Z):

  1. Velocity: vn=Ze22ϵ0nhv_n = \frac{Z e^2}{2 \epsilon_0 n h}

    This shows vnZnv_n \propto \frac{Z}{n}. For a given atom (Z is constant), vn1nv_n \propto \frac{1}{n}.

  2. Radius: rn=n2h2ϵ0πmeZe2r_n = \frac{n^2 h^2 \epsilon_0}{\pi m_e Z e^2}

    This shows rnn2Zr_n \propto \frac{n^2}{Z}. For a given atom (Z is constant), rnn2r_n \propto n^2.

Now, substitute these proportionalities into the frequency formula:

vn1/nn21n3v_n \propto \frac{1/n}{n^2} \propto \frac{1}{n^3}

So, the frequency of revolution vnv_n is inversely proportional to the cube of the principal quantum number nn. We can write this as:

vn=K1n3v_n = K \frac{1}{n^3}

where K is a constant of proportionality.

For the first Bohr orbit (n=1), the frequency of revolution v1v_1 is:

v1=K113=Kv_1 = K \frac{1}{1^3} = K

Now, let's find the ratio vn/v1v_n / v_1:

vnv1=K/n3K=1n3\frac{v_n}{v_1} = \frac{K/n^3}{K} = \frac{1}{n^3}

We need to find the graph between logn\log n and log(vn/v1)\log (v_n/v_1). Let's take the logarithm of both sides of the ratio equation:

log(vnv1)=log(1n3)\log \left(\frac{v_n}{v_1}\right) = \log \left(\frac{1}{n^3}\right)

Using logarithm properties, log(1/x)=log(x)\log(1/x) = -\log(x) and log(xy)=ylog(x)\log(x^y) = y \log(x):

log(vnv1)=log(n3)\log \left(\frac{v_n}{v_1}\right) = \log (n^{-3})

log(vnv1)=3logn\log \left(\frac{v_n}{v_1}\right) = -3 \log n

Let Y=log(vnv1)Y = \log \left(\frac{v_n}{v_1}\right) and X=lognX = \log n.

The equation becomes:

Y=3XY = -3X

This is the equation of a straight line passing through the origin (since Y=0 when X=0) with a negative slope (-3).

Comparing this with the given graphs:

(A) Shows a straight line with a positive slope passing through the origin.

(B) Shows a curve.

(C) Shows a straight line with a negative slope passing through the origin.

(D) Shows a straight line with a positive slope not passing through the origin.

Therefore, graph (C) correctly represents the relationship between logn\log n and log(vn/v1)\log (v_n/v_1).