Ultraviolet light of wavelength -1 and -2-2-1-when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies E1and E2 respectively- The value of Planck-s constant can be found from the relation -

Question

Toppr Admin · Accepted Answer

Suppose the frequency of the photons are $ν_{1}$ and $ν_{2}$ ( $ν_{1}$ > $ν_{1}$ ). Where, $ν = c / λ$ .
According to the question:
$h ν_{1} - E_{0} = E_{1}$
$h ν_{2} - E_{0} = E_{2}$ , where, $E_{0}$ is the ionization energy of the hydrogen.
Solving this two equation we will get:

$h = \frac{(E_{1} - E_{2}) λ_{1} λ_{2}}{c (λ_{2} - λ_{1})}$ .

So, the answer is option (C).

Ultraviolet light of wavelength λ1 and λ2(λ2>λ1)when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies E1and E2 respectively. The value of Planck's constant can be found from the relation :

Suppose the frequency of the photons are ν1 and ν2 (ν1>ν1). Where, ν=c/λ.According to the question:hν1−E0=E1hν2−E0=E2 , where, E0 is the ionization energy of the hydrogen.Solving this two equation we will get:h=(E1−E2)λ1λ2c(λ2−λ1).So, the answer is option (C).

Ultraviolet light of wavelength $λ_{1}$ and $λ_{2} (λ_{2} > λ_{1})$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies $E_{1}$ and $E_{2}$ respectively. The value of Planck's constant can be found from the relation :