An alpha particle and a proton are accelerated through the same potential difference find the ratio

We know that, Charge on the proton = e Charge on the alpha particle = 2e Let ​mass of proton = m So, mass of alpha particle = 4m When a particle is of mass m and change q is accelerated by a potential V, then its de Broglie wavelength is given by `λ=h/sqrt(2mqV)` For proton: `λ_"proton"=h/sqrt(2meV)` For alpha particle: `λ_"alpha particle"=h/sqrt(2×4m×2e×V)` `⇒λ_"alpha particle"/λ_"proton"=(h/(2×4m×2e×V))/(h/(2meV))` `⇒λ_"alpha particle"/λ_"proton"=(2meV)/(2×4m×2e×V)` `=1/(2sqrt2)`

Learning Objectives By the end of this section, you will be able to: • Define electric potential and electric potential energy. • Describe the relationship between potential difference and electrical potential energy. • Explain electron volt and its usage in submicroscopic process. • Determine electric potential energy given potential difference and amount of charge. Figure 1. A charge accelerated by an electric field is analogous to a mass going down a hill. In both cases potential energy is converted to another form. Work is done by a force, but since this force is conservative, we can write W = –ΔPE. When a free positive charge q is accelerated by an electric field, such as shown in Figure 1, it is given kinetic energy. The process is analogous to an object being accelerated by a gravitational field. It is as if the charge is going down an electrical hill where its electric potential energy is converted to kinetic energy. Let us explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electric potential energy. The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken. This is exactly analogous to the gravitational force in the absence of dissipative forces such as friction. When a force is conservative, it is possible to define a potential energy associated with the force, and it is usually easier to deal with the potential energy (because it depends ...

Einstein’s photoelectric equation is given by, Characteristic property of photons: i) Energy of photon is directly proportional to the frequency. ii) Total energy and momentum of the system of two constituent particles remain constant in photon-electron collision. Three observed features of photoelectric effect are: i) When frequency of incident photon increases, kinetic energy of emitted electron increases. Kinetic energy does not have any effect on Intensity of radiations. ii) When intensity of incident light increases, the number of incident photons increases. The increase in intensity will increase the number of ejected electrons. That is, photocurrent will increase with increase of intensity. Frequency has no effect on photocurrent. iii) When the energy of incident photon is greater than work function, the photoelectron is immediately ejected. Thus, there is no time lag between incidence of light and emission of photoelectrons. Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from to . Derive the expressions for the threshold wavelength and work function for the metal surface. Einstein’s photoelectric equation is given as: where, h is the energy of the photon, , is the work function of the metal, is the maximum kinetic energy of the emitted photoel...

Read the following statements : (A) Volume of the nucleus is directly proportional to the mass number. (B) Volume of the nucleus is independent of mass number. (C) Density of the nucleus is directly proportional to the mass number. (D) Density of the nucleus is directly proportional to the cube root of the mass number. (E) Density of the nucleus is independent of the mass number. Choose the correct option from the following options.

Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. • Article • • 07 June 2023 Interchange reconnection as the source of the fast solar wind within coronal holes • ORCID: orcid.org/0000-0002-1989-3596 • ORCID: orcid.org/0000-0002-9150-1841 • ORCID: orcid.org/0000-0001-6077-4145 • • ORCID: orcid.org/0000-0002-6145-436X • • ORCID: orcid.org/0000-0001-9898-464X • ORCID: orcid.org/0000-0002-7572-4690 • ORCID: orcid.org/0000-0003-2409-3742 • • • ORCID: orcid.org/0000-0001-6160-1158 • ORCID: orcid.org/0000-0002-0978-8127 • • • … • Show authors Nature volume 618, pages 252–256 ( 2023) The fast solar wind that fills the heliosphere originates from deep within regions of open magnetic field on the Sun called ‘coronal holes’. The energy source responsible for accelerating the plasma is widely debated; however, there is evidence that it is ultimately magnetic in nature, with candidate mechanisms including wave heating Recent measurements from the NASA Parker Solar Probe (PSP) showed that the solar wind emerging from coronal holes is organized into ‘microstreams’ with an angular scale (5–10°) in the Carrington longitude On solar Encounter 10 (E10), PSP came within 12.3 solar radii ( R S) of the photosphere. ...

Concept: The wavelength of any charged particle due to its motion is called the de-Broglie wavelength. When a charged particle is accelerated in a potential difference the energy gained by the particle is given by: Energy (E)= q × V Where V is the potential difference and q is the charge. Now, The de-Broglie wavelengthof charge particle (λ d) is given by: \( \) \(= 2\surd 2:1\)

The correct option is B √ 8 As the proton and alpha particle accelerating by the same potential difference. Hence, their KE will be K E = q V ⇒ 1 2 p 2 m = q V ⇒ p = √ 2 m q V [ ∵ V = same for both ] ∴ p p p α = √ q p m p q α m α de-Broglie wavelength is given by, λ = h p ⇒ λ ∝ 1 p ∴ λ p λ α = p α p p = √ q α m α q p m p ∵ q α = + 2 e , q p = + e and m α = 4 m p ∴ λ p λ α = √ ( + 2 e + e × 4 m p m p ) = √ 8 Hence, ( B ) is the correct answer. Why this question? Key note: Keep remember charge of alpha particle, q α = + 2 e. Sometimes it is not given in problem.