Atoms

Atoms

Easy Questions

Modern Physics

Dual Nature of Radiation and Matter
Atoms
Nuclei
Semiconductor Electronics : Materials, Devices and Simple Circuits
  1. ​An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be :​
    (m is the mass of the electron, R, Rydberg constant and h Planck’s constant)








  1. View Hint View Answer Discuss in Forum

    For emission, the wave number of the radiation is given as

    1
    		= RZ2
    1
    		-
    1
    λn21n22

    R = Rydberg constant, Z = atomic number
    = R
    1
    		-
    1
    1252

    = R1 -
    1
    25

    ⇒
    1
    		= R
    24
    λ25

    linear momentum
    P =
    h
    		= h × R ×
    24
    λ25

    (de-Broglie hypothesis)
    ⇒ mv =
    24hR
    		⇒ V =
    24hR
    2525m

    Correct Option: A

    For emission, the wave number of the radiation is given as

    1
    		= RZ2
    1
    		-
    1
    λn21n22

    R = Rydberg constant, Z = atomic number
    = R
    1
    		-
    1
    1252

    = R1 -
    1
    25

    ⇒
    1
    		= R
    24
    λ25

    linear momentum
    P =
    h
    		= h × R ×
    24
    λ25

    (de-Broglie hypothesis)
    ⇒ mv =
    24hR
    		⇒ V =
    24hR
    2525m

  1. An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is








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    KEmax = 10eV
    ​φ = 2.75 eV ​
    Total incident energy
    ​E = φ + KEmax = 12.75 eV ​
    ∴ Energy is released when electron jumps from the excited state n to the ground state. ​ 
    E4 – E4 = {– 0.85 – (–13.6) ev} ​= 12.75 eV ​
    ∴ value of n = 4

    Correct Option: B

    KEmax = 10eV
    ​φ = 2.75 eV ​
    Total incident energy
    ​E = φ + KEmax = 12.75 eV ​
    ∴ Energy is released when electron jumps from the excited state n to the ground state. ​ 
    E4 – E4 = {– 0.85 – (–13.6) ev} ​= 12.75 eV ​
    ∴ value of n = 4

  1. The energy of a hydrogen atom in the ground state is – 13.6 eV. The energy of a He+ ion in the first excited state will be​​








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    ​Energy of a H-like atom in it's nth state is given by

    En = -Z2 ×
    13.6
    		eV
    n2

    For, first excited state of He+,  n = 2, Z = 2
    EHe+ = -
    4
    		× 13.6 = - 13.6 eV
    22

    Correct Option: A

    ​Energy of a H-like atom in it's nth state is given by

    En = -Z2 ×
    13.6
    		eV
    n2

    For, first excited state of He+,  n = 2, Z = 2
    EHe+ = -
    4
    		× 13.6 = - 13.6 eV
    22

  1. ​Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr’s atomic model?








  1. View Hint View Answer Discuss in Forum

    Obviously, difference of 11.1eV is not possible.

    Correct Option: B

    Obviously, difference of 11.1eV is not possible.

  1. Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v1 and v2 then we must have ​








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    By law of conservation of energy, ​
    K.Ef = K.Ei – excitation energy (ε)
    or

    Correct Option: B

    By law of conservation of energy, ​
    K.Ef = K.Ei – excitation energy (ε)
    or