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in Atomic structure by (45 points)
15. Photo electric emission is observed from a surface for frequencies \( v_{1} \) and \( v_{2} \) the KE in two cases are in ratio \( 1: K \), then the threshold frequency \( v_{0} \) is given by 1) \( \frac{v_{2}-v_{1}}{K-1} \) 2) \( \frac{K v_{1}-v_{2}}{K-1} \) 3) \( \frac{K v_{2}-v_{2}}{K-1} \) 4) \( \frac{v_{2}-v_{2}}{K} \)

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2 Answers

+1 vote
by (25 points)

Option B ) is correct 

+1 vote
by (541 points)

HELLO,
 

As per question let assume two cases. In first case frequency is \(\upsilon_1\) and kinetic energy of emitted electron is \(k_1 \). In second case frequency is \(\upsilon_2 \) and kinetic energy of emitted electron is \(k_2 \).

Let work function of the given surface is \(\phi_0\).

Now,

    According to energy conversion the kinectic energy of emitted electron is equals to difeerence of energy of a photon and work function.

Mathematically,        \(K.E = E - \phi_0\)

                       and, \(E = h\upsilon\)

Now,   \(k_1 = h\upsilon_1-\phi_0\)      ------------  i

           \(k_2=h\upsilon_2-\phi_0\)     ------------  ii

Now, dividing eqn i and ii we get:-

                   \(\frac{k_1}{k_2} = \frac{h\upsilon_1-\phi_0}{h\upsilon_2-\phi_0}\)

According to question  \(\frac{k1}{k2} = \frac{1}{k}\)

and \(\phi_0 = h\upsilon_0\)   where \(\upsilon_0 \) is threshold frequency.

Now,

Subsituting the relations we get,

                       \(\frac{1}{k} = \frac{h(\upsilon_1-\upsilon_0)}{h(\upsilon_2-\upsilon_0)}\)

                       \(\upsilon_2 - \upsilon_0 = k\upsilon_1-k\upsilon_0\)

                       \(\upsilon_0(k-1) = k\upsilon_1 - \upsilon_2\)

                                \(\upsilon_0 = \frac{k\upsilon_1 - \upsilon_2}{k-1}\)

So, the correct option is B.

I HOPE YOU WILL UNDERSTAND.smiley 

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