Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainChemistryStructure of Atom

A proton and an alpha particle are accelerated from rest by different potential differences such that their final de Broglie wavelengths are identical. If the ratio of the accelerating potential of the proton to that of the alpha particle is x , then the value of x is _ _ _ _ _ .

Correct answer

8

Step-by-step solution

The de Broglie wavelength of a charged particle accelerated through a potential difference V is given by = h 2mqV . Given that _p = _ , we have: h 2m_p q_p V_p = h 2m_ q_ V_ Squaring both sides and rearranging gives: m_p q_p V_p = m_ q_ V_ For a proton, let mass m_p = m and charge q_p = e . For an alpha particle, mass m_ = 4m and charge q_ = 2e . Substituting these values: (m)(e)V_p = (4m)(2e)V_ V_p = 8 V_ The ratio V_p V_ = 8 . Answer: 8

Practice Structure of Atom on Quantrex Academy →

More from Structure of Atom

The 2s and the 2p orbital energies of hydrogen atom are E_ 2s ( H ) and E_ 2p ( H ) , respectively. The 2s and the 2p orbital energies of lithium atom are E_ 2s ( Li ) and E_ 2p ( 2026X ^ a+ and Y ^ b+ are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers n = 1 and n = 2 of X ^ a+ is . 2026Consider two radiations of wavelengths 1. ₁ = 2000 Å 2. ₂ = 6000 Å The ratio of the energies of these two radiations ( E₁ E₂ ) is ________ (Nearest integer). 2026The Bohr radius of a hydrogen like species is 70.53 pm. The species and the stationary state (n) are respectively (Given : Hydrogen atom Bohr radius is 52.9 pm) 2026If shortest wavelength of hydrogen atom in Lyman series is x , then longest wavelength in Balmer series of He ^+ is: 2026Match the List-I with List-II List-I Orbital List-II Radial nodes and nodal plane A. 2s I. 1 Radial node + two nodal planes B. 3s II. 1 Radial node + one nodal plane C. 3p III. 2 R 2026Which of the following statement(s) is/are true ? A. If two orbitals have the same value of (n + l) , the orbital with lower value of n will have lower energy. B. Energies of the o 2026Arrange the following atomic orbitals of multi electron atoms in order of increasing energy. A. n = 3, l = 2, m = +1 B. n = 4, l = 0, m = 0 C. n = 6, l = 1, m = 0 D. n = 5, l = 1, 2026 Full Structure of Atom list All JEE Main PYQs