Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the speeds are nonr

Question

Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the speeds are nonrelativistic.

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Acacia 3 years 2021-08-30T13:31:01+00:00 2 Answers 66 views 0

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  1. giabao
    0
    2021-08-30T13:32:13+00:00
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    Answer:

    the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal is 1835.16 .

    Explanation:

    We know, wavelength is expressed in terms of Kinetic Energy by :

    \lambda=\dfrac{h}{\sqrt{2mE}}

    Therefore , E=\dfrac{h^2}{2 \lambda^2 m}

    It is given that both electron and proton have same wavelength.

    Therefore,

    E_e=\dfrac{h^2}{2 \lambda^2 m_e}   …. equation 1.

    E_p=\dfrac{h^2}{2 \lambda^2 m_p}   …. equation 2.

    Now, dividing equation 1 by 2 .

    We get ,

    \dfrac{E_e}{E_p}=\dfrac{\dfrac{h^2}{2 \lambda^2 m_e}}{\dfrac{h^2}{2 \lambda^2 m_p}}\\\\\\\dfrac{E_e}{E_p}=\dfrac{m_p}{m_e}

    Putting value of mass of electron = 9.1\times 10^{-31}\ kg and mass of proton = 1.67\times 10^{-27}\ kg.

    We get :

    \dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16

    Hence , this is the required solution.

  2. Verity
    0
    2021-08-30T13:32:26+00:00
    Reply

    Answer:

    \frac{KE_e}{KE_p}=1835.16

    Explanation:

    Given that the wavelengths of electron and proton are equal at non- relativistic speed.

    From De-Broglie wave equation we know that:

    \lambda =\frac{h}{p}

    where:

    \lambda= wavelength

    h= Planck’s constant

    p= linear momentum of  the particle

    Then’

    \lambda_e=\lambda_p

    \frac{h}{p_e} =\frac{h}{p_p}

    \frac{1}{m_e.v_e} =\frac{1}{m_p.v_p} …………………………….(1)

    we’ve mass of electron, m_e=9.1\times 10^{-31}\ kg

    mass pf proton, m_p=1.67\times 10^{-27}\ kg

    Now,

    kinetic energy of electron:

    KE_e=\frac{1}{2} m_e.v_e^2

    kinetic energy of proton:

    KE_p=\frac{1}{2}m_p.v_p^2

    So,

    \frac{KE_e}{KE_p}=\frac{m_e.v_e^2}{m_p.v_p^2}

    from eq. (1)

    \frac{KE_e}{KE_p}=\frac{m_e}{m_p} \times \frac{m_p^2}{m_e^2}

    \frac{KE_e}{KE_p}= \frac{m_p}{m_e}

    \frac{KE_e}{KE_p}=\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}

    \frac{KE_e}{KE_p}=1835.16

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