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Chapter 17: Problem 2407

The cathode of a photoelectric cell is changed such that the work function changes from \(\mathrm{W}_{1}\) to \(\mathrm{W}_{2}\left(\mathrm{~W}_{2}>\mathrm{W}_{1}\right)\). If the currents before and after change are \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\), all other conditions remaining unchanged, then assuming $\mathrm{hf}>\mathrm{W}_{2} \ldots \ldots$ (A) \(\mathrm{I}_{1}=\mathrm{I}_{2}\) (B) \(I_{1}\mathrm{I}_{2}\) (D) \(\mathrm{I}_{1}<\mathrm{I}_{2}<2 \mathrm{I}_{1}\)

Short Answer

Expert verified
The short answer to the given problem is, \(I_1 < I_2\).

Step by step solution

01

Understand the photoelectric effect

According to the photoelectric effect, when light of sufficient energy (hf) falls on a metal, electrons are emitted from the metal surface. The minimum energy (called work function, W) required to remove an electron from the metal surface should be less than or equal to the energy of the incoming photons (hf).
02

Apply the photoelectric equation to each cathode

The photoelectric equation for each cathode can be written as: Cathode 1: \(hf = W_1 + KE1\) Cathode 2: \(hf = W_2 + KE2\) Here, KE1 and KE2 represent the kinetic energy of the emitted electrons in each case.
03

Calculate kinetic energy in each case

Subtract the equations for Cathode 1 and Cathode 2 to find the difference in kinetic energies of the emitted electrons: \(W_2 - W_1 = KE1 - KE2\) Since \(W_2 > W_1\), it follows that \(KE1 < KE2\).
04

Understand the relation between current and emitted electrons

Now, we have to relate the kinetic energies of the emitted electrons to the currents I1 and I2. In general, the number of emitted electrons determines the current in a photoelectric cell. For a given energy of light (hf), if the kinetic energy of emitted electrons is higher, they are more likely to overcome the potential barrier and create a current.
05

Determine the relationship between I1 and I2

Since the kinetic energy of emitted electrons in case of Cathode 2 is higher, it means more electrons will contribute to the current, i.e., \(I_2 > I_1\). This matches option (C): \(I_1 < I_2\). Hence, the answer is (C): \(I_1 < I_2\).

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Most popular questions from this chapter

In an experiment to determine photoelectric characteristics for a metal the intensity of radiation is kept constant. Starting with threshold frequency. Now, frequency of incident radiation is increased. It is observed that $\ldots \ldots \ldots$ (A) the number of photoelectrons increases (B) the energy of photoelectrons decreases (C) the number of photoelectrons decreases (D) the energy of photoelectrons increases.

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

Work function of tungsten and sodium are \(4.5 \mathrm{eV}\) and $2.3 \mathrm{eV}\( respectively. If threshold wavelength for sodium is \)5460 \mathrm{~A}$, threshold frequency for tungsten will be ............ (A) \(10^{15} \mathrm{~Hz}\) (B) \(1.1 \times 10^{15} \mathrm{~Hz}\) (C) \(1.2 \times 10^{15} \mathrm{~Hz}\) (D) \(1.4 \times 10^{15} \mathrm{~Hz}\)

When a radiation of wavelength \(3000 \AA\) is incident on metal, $1.85 \mathrm{~V}$ stopping potential is obtained. What will be threshed wave length of metal? $\left\\{\mathrm{h}=66 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(4539 \AA\) (B) \(3954 \AA\) (C) \(5439 \AA\) (D) \(4395 \AA\)

A photon having energy \(5.5 \mathrm{eV}\) is incident on metal suface and emits photo electrons having maximum kinetic energy \(0.4 \mathrm{eV}\). Then stopping potential of this electron is..... (A) \(5.5 \mathrm{eV}\) (B) \(5.1 \mathrm{eV}\) (C) \(5.9 \mathrm{eV}\) (D) \(0.4 \mathrm{eV}\)
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