\(\frac{hc}{E\lambda - hc}\)
\(\frac{hc\lambda}{E\lambda + hc}\)
\(\frac{h\lambda}{E\lambda + hc}\)
\(\frac{hc\lambda}{E\lambda - hc}\)
To determine the wavelength required to double the kinetic energy of the emitted electron, we need to apply the concept of photoelectric effect. According to Einstein's photoelectric equation, the kinetic energy \( K \) of the emitted electron when light of wavelength \( \lambda \) is incident on a metal surface is given by:
\(K = \frac{hc}{\lambda} - \phi\)
where:
Given that the initial kinetic energy \( E \) can be expressed as:
\(E = \frac{hc}{\lambda} - \phi\)
We want to double this kinetic energy, so we have:
\(2E = \frac{hc}{\lambda'} - \phi\)
where \( \lambda' \) is the new wavelength. We can set up the equation for \( 2E \) and equate it to the adjusted photoelectric equation:
\(2E = \frac{hc}{\lambda'} - \left( \frac{hc}{\lambda} - E \right)\)
Simplifying this equation:
\(2E = \frac{hc}{\lambda'} + E - \frac{hc}{\lambda}\)
\(2E - E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}\)
\(E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}\)
Now solve for \( \lambda' \):
\(\frac{hc}{\lambda'} = E + \frac{hc}{\lambda}\)
\(\lambda' = \frac{hc}{E + \frac{hc}{\lambda}} = \frac{hc \cdot \lambda}{E \lambda + hc}\)
Thus, the new wavelength required to double the kinetic energy is:
\(\lambda' = \frac{hc \lambda}{E \lambda + hc}\)
This matches the provided option:
\(\frac{hc\lambda}{E\lambda + hc}\)
Therefore, the correct answer is indeed \(\frac{hc\lambda}{E\lambda + hc}\), which is consistent with the principles of the photoelectric effect.
\[ k = \frac{hc}{\lambda} - \Phi = E \] and,
\[ 2k = \frac{hc}{\lambda^2} - \Phi = 2E \]
\[ \Rightarrow \frac{hc}{\lambda} - E = \frac{hc}{\lambda^2} - 2E \]
\[ \Rightarrow \frac{hc}{\lambda^2} = \frac{hc}{\lambda} + E \]
\[ \Rightarrow \lambda^2 = \frac{hc\lambda}{hc + \lambda E} \]
So, the correct option is (B).
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
According to Einstein’s explanation of the photoelectric effect :
The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron
i.e. hν = W + E
Where,