/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 The energy that should be added ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The energy that should be added to an electron, to reduce its de-Broglie wavelengths from \(10^{-10} \mathrm{~m}\) to \(0.5 \times 10^{-10} \mathrm{~m}\), will be (A) four times the initial energy. (B) thrice the initial energy. (C) equal to the initial energy. (D) twice the initial energy.

Short Answer

Expert verified
To find the energy that should be added to the electron, we use the de-Broglie wavelength formula, the momentum formula, and the kinetic energy formula. First, we compute the initial and final kinetic energy using the given wavelengths: \(K_{initial}\) from \(\lambda_{initial} = 10^{-10} \mathrm{~m}\) and \(K_{final}\) from \(\lambda_{final} = 0.5 \times 10^{-10} \mathrm{~m}\). Then, we find the energy difference, \(\Delta K = K_{final} - K_{initial}\). Finally, we compare the added energy with the initial energy to select the correct option.

Step by step solution

01

Write down the de-Broglie wavelength formula

The de-Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the de-Broglie wavelength, \(h\) is the Planck's constant (\(h = 6.626 \times 10^{-34} \text{ Js}\)), and \(p\) is the momentum of the particle.
02

Write down the formula for momentum and kinetic energy

The momentum formula is given by: \[ p = \sqrt{2\cdot m_e \cdot K} \] where \(p\) is the momentum, \(m_e\) is the electron's mass (\(m_e = 9.109\times 10^{-31}\, \text{kg}\)), and \(K\) is the kinetic energy of the electron.
03

Find the initial kinetic energy using the initial wavelength

We first find the initial momentum using the initial de-Broglie wavelength: \[ p_{initial} = \frac{h}{\lambda_{initial}} \] Now, we can find the initial kinetic energy (K_i) using the momentum formula: \[ K_{initial} = \frac{p_{initial}^2}{2\cdot m_e} \]
04

Find the final kinetic energy using the final wavelength

Similarly, we can find the final momentum using the final de-Broglie wavelength: \[ p_{final} = \frac{h}{\lambda_{final}} \] Then, we can find the final kinetic energy (K_f) using the momentum formula: \[ K_{final} = \frac{p_{final}^2}{2\cdot m_e} \]
05

Calculate the energy that should be added to the electron

The energy that should be added to the electron is given by the difference between the final kinetic energy and the initial kinetic energy: \[ \Delta K = K_{final} - K_{initial} \]
06

Compare the added energy with the initial energy

Now, we compare the energy that should be added to the electron with the initial energy to find which one of the options is correct. For example, if the added energy is equal to twice the initial energy, then the correct option would be (D) twice the initial energy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An alpha nucleus of energy \(\frac{1}{2} m v^{2}\) bombards a heavy nuclear target of charge \(\mathrm{Ze}\). Then the distance of closest approach for the alpha nucleus will be proportional to (A) \(v^{2}\) (B) \(\frac{1}{m}\) (C) \(\frac{1}{v^{2}}\) (D) \(\frac{1}{Z e}\)

The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number \(n\) are \(E, P\) and \(r\) respectively. Then according to Bohr's theory for hydrogen atom (A) \(E P r\) is proportional to \(\frac{1}{n}\). (B) \(\frac{P}{E}\) is proportional to \(n\). (C) \(E r\) is constant for all orbits. (D) \(P r\) is proportional to \(n\).

A radioactive nucleus (initial mass number \(A\) and atomic number \(Z\) ) emits \(3 \alpha\)-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be (A) \(\frac{A-Z-8}{Z-4}\) (B) \(\frac{A-Z-4}{Z-8}\) (C) \(\frac{A-Z-12}{Z-4}\) (D) \(\frac{A-Z-4}{Z-2}\)

The transition from the state \(n=4\) to \(n=3\) in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from: (A) \(3 \rightarrow 2\) (B) \(4 \rightarrow 2\) (C) \(5 \rightarrow 4\) (D) \(2 \rightarrow 1\)

Which of the following statement about \(x\)-rays is/are true? (A) \(E\left(K_{\alpha}\right)+E\left(L_{\beta}\right)=E\left(K_{\beta}\right)+E\left(M_{\alpha}\right)=E\left(K_{\gamma}\right)\), where \(E\) is the energy of respective \(x\)-rays. (B) For the harder \(x\)-rays, the intensity is higher than soft \(x\)-rays. (C) The continuous and the characteristic \(x\)-rays differ only in the method of creation. (D) The cut-off wavelength \(\lambda_{\min }\) depends only on the accelerating voltage applied between the target and the filament.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.