$\text{A photon and an electron (mass m) have the same energy E. The ratio (λphoton/λelectron) of their de Broglie wavelengths is: (c is the speed of light)}$
$\text{(1) c√(2mE)}$ $\text{ (2) c√(2m/E)}$
$\text{(3) (1/c)√(E/2m)}$ $\text{ (4) √(E/2m)}$
$\text{Answer (2)}$

The correct option is Option (2): c√(2m/E).

To find the ratio of the de Broglie wavelength of a photon to that of an electron, we can derive the expressions for both wavelengths when they have the same energy $E$.

Step 1: Find the wavelength of the photon (${\lambda }_{\text{photon}}$)
The energy $E$ of a photon is related to its wavelength ${\lambda }_{\text{photon}}$ by the relation:
$E=\frac{hc}{{\lambda }_{\text{photon}}}$
where $h$ is Planck's constant and $c$ is the speed of light.
Rearranging the equation for ${\lambda }_{\text{photon}}$, we get:
${\lambda }_{\text{photon}}=\frac{hc}{E}$           (Equation 1)

Step 2: Find the de Broglie wavelength of the electron (${\lambda }_{\text{electron}}$)
The de Broglie wavelength of a particle with mass $m$ and momentum $p$ is given by:
${\lambda }_{\text{electron}}=\frac{h}{p}$
The kinetic energy $E$ of the electron is related to its momentum $p$ by:
$E=\frac{p^2}{2m}\Rightarrow p=\sqrt{2mE}$
Substituting this momentum into the de Broglie wavelength formula:
${\lambda }_{\text{electron}}=\frac{h}{\sqrt{2mE}}$           (Equation 2)

Step 3: Calculate the ratio ($\frac{{\lambda }_{\text{photon}}}{{\lambda }_{\text{electron}}}$)
Dividing Equation 1 by Equation 2:
$\frac{{\lambda }_{\text{photon}}}{{\lambda }_{\text{electron}}}=\frac{\left(\frac{hc}{E}\right)}{\left(\frac{h}{\sqrt{2mE}}\right)}$
Simplifying the expression:
$\frac{{\lambda }_{\text{photon}}}{{\lambda }_{\text{electron}}}=\frac{hc}{E}·\frac{\sqrt{2mE}}{h}$
$\frac{{\lambda }_{\text{photon}}}{{\lambda }_{\text{electron}}}=\frac{c\sqrt{2mE}}{E}$
We can write $E$ in the denominator as $\sqrt{E^2}$ to bring it inside the square root:
$\frac{{\lambda }_{\text{photon}}}{{\lambda }_{\text{electron}}}=c\sqrt{\frac{2mE}{E^2}}$
$\frac{{\lambda }_{\text{photon}}}{{\lambda }_{\text{electron}}}=c\sqrt{\frac{2m}{E}}$

Thus, the ratio of their wavelengths is indeed $c\sqrt{\frac{2m}{E}}$.