A radiation of wavelength  λ  illuminates a metal and ejects photoelectrons of maximum kinetic energy of  1eV . Another radiation of wavelength  λ3  ejects photoelectrons of maximum kinetic energy of  4eV . What will be the work function of metal in terms of  x × 10−1eV , where  x  is_________.

The photoelectric effect can be described by the equation:

E = hf - Φ

where E is the energy of the ejected electron, h is Planck's constant, f is the frequency of the incident light, and Φ is the work function of the metal.

We can rewrite this equation in terms of wavelength (λ) as:

E = hc/λ - Φ

where c is the speed of light.

Given that the energy E is given in electronvolts (eV), we can use the values h = 4.1357 x 10^-15 eV*s and c = 2.998 x 10^8 m/s.

From the problem, we know that for a wavelength λ, the energy E is 1 eV. So we can write:

1 eV = (4.1357 x 10^-15 eV*s * 2.998 x 10^8 m/s) / λ - Φ

Solving for Φ gives us:

Φ = (4.1357 x 10^-15 eV*s * 2.998 x 10^8 m/s) / λ - 1 eV

Similarly, for a wavelength λ3, the energy E is 4 eV. So we can write:

4 eV = (4.1357 x 10^-15 eV*s * 2.998 x 10^8 m/s) / λ3 - Φ

Solving for Φ gives us:

Φ = (4.1357 x 10^-15 eV*s * 2.998 x 10^8 m/s) / λ3 - 4 eV

Since the work function Φ is the same for both wavelengths, we can set these two equations equal to each other and solve for λ3:

(4.1357 x 10^-15 eVs * 2.998 x 10^8 m/s) / λ - 1 eV = (4.1357 x 10^-15 eVs * 2.998 x 10^8 m/s) / λ3 - 4 eV

Solving this equation for λ3 gives us:

λ3 = 4λ

Substituting this back into the equation for Φ gives us:

Φ = (4.1357 x 10^-15 eV*s * 2.998 x 10^8 m/s) / (4λ) - 4 eV

Solving for Φ gives us:

Φ = 1 eV

So the work function of the metal is 1 eV, or 10^-1 eV. Therefore, x = 1.