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Chapter 18: Problem 2520

If a hydrogen atom emits a Photon of wave length \(\lambda\). the recoil speed of the atom of mass \(\mathrm{m}\) is given by (A) \((\mathrm{h} / \mathrm{m} \lambda)\) (B) \((\mathrm{mh} / \lambda)\) (C) \(\operatorname{mh} \lambda\) (D) \((\mathrm{m} \lambda / \mathrm{h})\)

Short Answer

Expert verified
The short answer is: The recoil speed of the hydrogen atom is given by v = \(\frac{h}{m \lambda}\), which matches option (A).

Step by step solution

01

Understand the given parameters and variables

We are given the following variables and parameters: - Wavelength of the emitted photon: \(\lambda\) - Mass of the hydrogen atom: m - Planck's constant: h Our task is to find the correct expression for the recoil speed of the hydrogen atom, given these parameters.
02

Find the momentum of the emitted photon

According to the de Broglie hypothesis, the momentum of a photon can be expressed as: Momentum of the photon (p_photon) = \(\frac{h}{\lambda}\) Where h is the Planck's constant and \(\lambda\) is the wavelength of the emitted photon.
03

Apply the conservation of momentum to find the momentum of the hydrogen atom

According to conservation of momentum, the momentum of the hydrogen atom after the emission of the photon is equal and opposite to the momentum of the emitted photon. Thus, the momentum of the hydrogen atom can be expressed as: Momentum of the hydrogen atom (p_hydrogen) = -p_photon
04

Find the recoil speed of the hydrogen atom

The momentum of the hydrogen atom can be related to its mass, m, and its recoil speed, v, by the following expression: p_hydrogen = m * v We can now substitute the expression for p_hydrogen from Step 3: m * v = -\(\frac{h}{\lambda}\) Now, we need to find the expression for the recoil speed, v. We can do this by isolating v: v = -\(\frac{h}{m \lambda}\) Considering that only the magnitudes are shown in the options, we can disregard the negative sign: v = \(\frac{h}{m \lambda}\) Comparing this expression to the provided options, we can see that it matches option (A). Thus, the recoil speed of the hydrogen atom is given by: v = \(\frac{h}{m \lambda}\)

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Most popular questions from this chapter

Read the following question and choose correct Answer form given below. (A) Both assertion and reason are true. Reason is the correct explanation of the Assertion. (B) Both assertion and reason are true. Reason is not correct explanation of the assertion. (C) Assertion is true but reason is false. (D) Assertion is false and Reason are true. (i) Assertion :- In a radio-active disintegration, an electron is emitted by nucleus. Reason :- electron are always Present in-side the nucleus. (ii) Assertion :- An electron and Positron can annihilate each other creating Photon Reason:- Electron and Positron form a Particle and anti Particle pair. (iii) Assertion:- An isolated radioactive atom may not decay at all what ever be its half time Reason:- Radioactive decay is a statistical Phenomena. (iv) Assertion:- Fragment Produced in the fission of \(\mathrm{u}^{235}\) are active Reason:- The fragments have abnormally high Proton to neutron ratio

If the binding energy of electron in a hydrogen atom is \(13.6 \mathrm{eV}\), the energy required to remove the electron form the first state of \(\mathrm{Li}^{2+}\) is. (A) \(13.6 \mathrm{eV}\) (B) \(30.6 \mathrm{eV}\) (C) \(122.4 \mathrm{eV}\) (D) \(3.4 \mathrm{eV}\)

The hydrogen atom can give spectral lines in the series Lyman, Balmer and Paschen. which of the following statement is correct (A) Lyman series is in the infrared region (B) Balmer series is in the ultraviolet region (C) Balmer series is in the visible region (D) Paschen series is in the visible region.

excited hydrogen atom emits a Photon of wave length \(\lambda\) in returning to the ground state The quantum number \(\mathrm{n}\) of excited state is given by (A) \(\sqrt{[}(\lambda . \mathrm{R}-1) /(\lambda \mathrm{R})]\) (B) \(\sqrt{[}(\lambda \mathrm{R}) /(\lambda \mathrm{R}-1)]\) (C) \(\sqrt{[\lambda R}(\lambda \mathrm{R}-1)]\) (D) \(\lambda \mathrm{R}(\mathrm{R}-1)\)

A radiation of energy \(\mathrm{E}\) falls normally on a Perfect reflecting surface. The momentum transferred to the surface is. (A) \((\mathrm{E} / \mathrm{c})\) (B) \((2 \mathrm{E} / \mathrm{c})\) (C) \(\left(\mathrm{E} / \mathrm{c}^{2}\right)\) (D) Ec
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