Photons with energy 5, eV are incident on a c | vedclass

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

Question

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by:$KE_{\max} = E - \phi$For the first case,$

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy of photoelectrons is given by:$KE_{\max} = E - \phi$For the first case,$E = 5\, eV$ and $KE_{\max} = 2\, eV$:$2 = 5 - \phi \implies \phi = 3\, eV$When photons of energy $E' = 6\, eV$ are incident,the new maximum kinetic energy is:$KE'_{\max} = E' - \phi = 6 - 3 = 3\, eV$For no photoelectrons to reach the anode $A,$ the potential of the anode relative to the cathode $(V_A - V_C)$ must be equal to the negative of the stopping potential $(V_s)$.The stopping potential $V_s$ is defined by $e V_s = KE'_{\max} = 3\, eV,$ so $V_s = 3\, V$.Since the anode must be at a negative potential relative to the cathode to stop the electrons,the potential difference $V_A - V_C = -3\, V$.

Similar Questions

The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

When two monochromatic lights of frequency $v$ and $\frac{v}{2}$ are incident on a photoelectric metal,their stopping potentials are $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is:

$A$ point source of light is used in a photoelectric effect. If the source is moved farther from the emitting metal, the stopping potential:

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is $6250 \ \text{Å}$, then the work function of the metal is (Planck's constant $= 6.6 \times 10^{-34} \ \text{Js}$) (in $\text{eV}$)

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} \, m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} \, m$ is (Given $h = 4.14 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module