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Photoelectric Emission Dual Nature of Radiation and Matter Physics Class 12th

In a photoelectric experiment, increasing the intensity of incident light:

Option 1 -

Increases the number of photons incident and also increases the K.E. of the ejected electrons.

Option 2 -

Increases the number of photons incident and the K.E. of the ejected electrons remains unchanged.

Option 3 -

Increases the frequency of photons incident and increases the K.E. of the ejected electrons.

Option 4 -

Increases the frequency of photons incident and the K.E. of the ejected electrons remains unchanged

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 weeks ago

Correct Option - 3

Detailed Solution:

Intersity a number of photons kinetic Energy a f

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The work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron whenThe work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron when\

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For no emission of electron

λ > λ0

Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is

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According to Einstein’s photoelectric equation maximum kinetic energy of photoelectrons, $K E_{max} = E_{v} - φ$

or $2 = 5 - φ$ $∴ φ = 3 e V$

When $E_{v} = 6 e V$ then, $K E_{max} = 6 - 3 = 3 e V$

or $e (V_{c a t h o d e} - V_{a n o d e}) = 3 e V$

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The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are $2.14 e V, 2.30 e V$ and $2.75 e V$ respectively. If incident electromagnetic radiation has an incident energy of $2.20 e V$ , which of these photosensitive surfaces may emit photoelectrons?

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Incident energy $= 2.20 e V$

If $? < 2.20 e V$ electron will emit.

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Only caesium will emit electron

Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant $^{}$ )

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$e V s = h v - ?$

$\begin{array}{r} At V_{s} = 0 \Rightarrow h v = ? \\ \Rightarrow ? = [6.62 \times 10^{- 34}] [10^{14}] [5.5] \\ \Rightarrow ? = \frac{ [6.62 \times 10^{- 34}] [10^{14}] [5.5] e V}{(1.6 \times 10^{- 19}]} \end{array}$

$= 2.27$

When photon of energy $4.0 e V$ strikes the surface of a metal $A$ , the ejected photoelectrons have maximum kinetic energy $T_{A}$ $_{}$ eV and de-Broglie wavelength $λ_{A}$ $_{}$ . The maximum kinetic energy of photoelectrons liberated form another metal B by photon of energy 4.50 $e V$ is $T_{B} = (T_{A} - 1.5) e V$ $_{} (_{})$ . If the de-Broglie wavelength of these photoelectrons $λ_{B} = 2 λ_{A}$ $_{}_{}$ , then the work function of metal $B$ is

Raj Pandey

Relation between De-Broglie wavelength and K.E. is

$λ = \frac{h}{\sqrt[]{2 (K E) m_{e}}} \Rightarrow λ \propto \frac{1}{\sqrt[]{K E}}$

$\begin{array}{r} \frac{λ_{A}}{λ_{B}} = \frac{\sqrt[]{{K E}_{B}}}{\sqrt[]{K E}} \Rightarrow \frac{1}{2} = \sqrt[]{\frac{T_{A} - 1.5}{T_{A}}} \\ \Rightarrow T_{A} 2 e V \\ ∴ {K E}_{B} = 2 - 1.5 = 0.5 e V \\ ?_{B} = 4.5 - 0.5 = 4 e V \end{array}$

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