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**Hint:**We all know that to eject an electron from the metal surface, we need a specific amount of energy with a particular frequency and that frequency is called threshold frequency which when provided to the electrons so that they can come out of the surface.

**Complete step by step answer:**

As we all know the that by the Einstein’s photoelectric equation, the energy of the photon is given by:

$E = W + KE$ …… (I)

Here E is the energy of the photon , W is the work potential and KE is the kinetic energy of the photo electrons.

As we all know that by Einstein-Planck's equation:

$E = \dfrac{{hc}}{\lambda }$

Here h is the planck’s constant, c is the speed of light and $\lambda $ is the wavelength of emitted radiation.

We will substitute $E = \dfrac{{hc}}{\lambda }$ in equation (I) to finalise the equation. Hence we will get,

$ \Rightarrow \dfrac{{hc}}{\lambda } = W + KE$ ……. (II)

So for the state 1, we can write the equation (II) as, the subscript 1 tells about state 1.

\[ \Rightarrow \dfrac{{hc}}{\lambda } = W + K{E_1}\] …… (III)

Now since the kinetic energy is doubled , so the equation (II) becomes,

$ \Rightarrow \dfrac{{hc}}{{\lambda '}} = W + 2K{E_1}$ …… (IV)

Now we will divide equation (I) from equation (II) and we will get,

\[

\Rightarrow \dfrac{{\dfrac{{hc}}{\lambda }}}{{\dfrac{{hc}}{{\lambda '}}}} = \dfrac{{W + K{E_1}}}{{W + 2K{E_1}}} \\

\Rightarrow \dfrac{{\lambda '}}{\lambda } = \dfrac{{W + K{E_1}}}{{W + 2K{E_1}}} \\

\therefore \dfrac{{\lambda '}}{\lambda } < 1 \\

\]

Now we can clearly say that, \[\lambda ' < \lambda \]. Therefore, the only option matches is (C). Hence the correct option is (C).

**Note:**As we all know that Einstein produced a very effective model of radiation and told that light consists of a very small number of particles. These are not the matter particles but are the packets of pure energy. Each of the packets are called the quantum of energy and are called photons.

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