The stopping potential for the photoelectrons | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$(A)$ According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = e V_{0} = E - \p

Vedclass · Accepted Answer

$n = 3$ to $n = 1$

Vedclass · Answer

$(A)$ According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = e V_{0} = E - \phi_{0}$, where $E$ is the energy of the incident photon, $V_{0}$ is the stopping potential, and $\phi_{0}$ is the work function.Given: $V_{0} = 10.4 \ V$ and $\phi_{0} = 1.7 \ eV$.Thus, $E = e V_{0} + \phi_{0} = 10.4 \ eV + 1.7 \ eV = 12.1 \ eV$.The energy of a photon emitted during a transition in a hydrogen atom from energy level $n_{i}$ to $n_{f}$ is given by $\Delta E = 13.6 \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} ight) \ eV$.For transition from $n = 3$ to $n = 1$:$\Delta E = 13.6 \left( \frac{1}{1^{2}} - \frac{1}{3^{2}} ight) = 13.6 \left( 1 - \frac{1}{9} ight) = 13.6 	imes \frac{8}{9} \approx 12.09 \ eV \approx 12.1 \ eV$.Therefore, the transition from $n = 3$ to $n = 1$ corresponds to the energy of the incident photons.

Similar Questions

The wavelength of the first spectral line of the Lyman series of the hydrogen spectrum is

The ratio of minimum wavelength of Balmer series to maximum wavelength in Brackett series in hydrogen spectrum is

The minimum wavelength of Lyman series lines is $P$. What is the maximum wavelength of these lines?

Which of the following is true for the number of spectral lines when going from the Lyman series to the Pfund series?

$A$ hydrogen atom in the ground state is given an energy of $10.2 \ eV$. How many spectral lines will be emitted due to the transition of electrons?

Vedclass Test Series

Exam Paper Generator

Online Exam Module