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According to Einstein's photoelectric equation, graph of kinetic energy of emitted photo electrons from metal versus frequency of incident radiation is linear. Its slope.......... (A) depends on type of metal used (B) depends on intensity of radiation (C) depends on both metal used and intensity of radiation. (D) is same for all metals and free from intensity of radiation.

Short Answer

Expert verified
The correct answer is (D). The slope of the graph of kinetic energy of emitted photoelectrons from metal versus frequency of incident radiation is the same for all metals and free from intensity of radiation, as it depends only on Planck's constant \(h\), which is universal.

Step by step solution

01

Understand Einstein's Photoelectric Equation

Einstein's photoelectric equation is given by: \[ KE = h \nu - \phi \] Here, - \(KE\) represents the kinetic energy of emitted photoelectrons, - \(h\) is Planck's constant, - \(\nu\) is the frequency of incident radiation, and - \(\phi\) is the work function of the metal (energy required to remove an electron from the metal) From this equation, we can see that it follows a straight line equation (y = mx + c). To establish a correlation with the given factors, we will rewrite this equation in the form of a linear equation with \(KE\) as the dependent variable and \(\nu\) as the independent variable.
02

Rewrite the Equation in Linear Form

Rewrite the photoelectric equation as a linear equation: \- KE = h⋅ν - φ We can rewrite the equation of 'y = mx + c' as follows: \[KE = (h) \nu - \phi \] In this linear form, the slope of the graph is given by the coefficient of the independent variable (\(h\)):
03

Analyze the Slope and its Factors

From the linear form of the equation, we can see the slope of the graph of the kinetic energy of emitted photoelectrons from metal versus the frequency of incident radiation is given by Planck's constant (\(h\)). Planck's constant is universal and does not depend on the type of metal nor the intensity of radiation. Now let's analyze the given options: (A) depends on the type of metal used: *False*, as the slope depends on Planck's constant \(h\) which is universal. (B) depends on the intensity of radiation: *False*, as Planck's constant \(h\) is universal and not related to the intensity of radiation. (C) depends on both metal used and intensity of radiation: *False*, as the slope only depends on Planck's constant \(h\). (D) is the same for all metals and free from intensity of radiation: *True*, as the slope only depends on Planck's constant \(h\), which is universal and does not depend on the type of metal or the intensity of radiation.
04

Conclusion

Based on our analysis, we can conclude that the correct answer is (D), the slope of the graph of the kinetic energy of emitted photoelectrons from metal versus the frequency of incident radiation is the same for all metals and free from intensity of radiation.

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Most popular questions from this chapter

A proton and an \(\propto\) -particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is \(\ldots \ldots\) (A) \(1: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(2 \sqrt{2}: 1\)

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Energy of photon of wavelength \(\lambda\) is (P) \((\mathrm{E} / \mathrm{p})\) (II) The de Broglie wavelength associated (Q) \(\left(\mathrm{hf} / \mathrm{c}^{2}\right)\) with particle of momentum \(\mathrm{P}\) is (II) Mass of photon in motion is (R) (hc \(/ \lambda\) ) (IV) The velocity of photon of energy (S) \((\mathrm{h} / \mathrm{p})\) \(\mathrm{E}\) and momentum \(\mathrm{P}\) is (A) I - P, II - Q. III - R, IV - S (B) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$ (C) $\mathrm{I}-\mathrm{R}, \mathrm{II}-\mathrm{S}, \mathrm{III}-\mathrm{P}_{3} \mathrm{IV}-\mathrm{Q}$ (D) $\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{P}$

If de-Broglie wavelength of electron is increased by \(1 \%\) its momentum \(\ldots \ldots\) (A) increases by \(1 \%\) (B) decreases by \(1 \%\) (C) increased by \(2 \%\) (D) decreases by \(2 \%\)

Compare energy of a photon of X-rays having \(1 \AA\) wavelength with the energy of an electron having same de Broglie wavelength. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(8.24\) (B) \(2.48\) (C) \(82.4\) (D) \(24.8\)

The de-Broglie wavelength of a proton and \(\alpha\) - particle is same. The ratio of their velocities will be.......... ( \(\alpha\) particle is the He- nucleus, having two protons and two neutrons. Thus, its mass \(\mathrm{M}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\) where \(\mathrm{m}_{\mathrm{p}}\) is the mass of the proton.) (A) \(1: 4\) (B) \(1: 2\) (C) \(2: 1\) (D) \(4: 1\)

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