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If the binding energy of electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron form the first state of $Li^{2+}$ is.
Binding energy = ${13.6 Z^2 \over n^2}$ For $Li^{2+}$ Z = 3, n = 2 first exited state
The ionization Potential of hydrogen atom is 13.6 eV. An electron in the ground state absorbs Photon of energy 12.75 eV. How many different spectral lines can one expect when electron make a down ward transition
$\triangle E = E_n - E_1$ $E_n = \triangle E + E_1 = 12.75-13.6 = -0.85$ $E_n = -{13.6 \over n^2}$ $ -0.85 = {-13.6 \over n^2} $ $ n^2 = {13.6 \over 0.85} = 16 --> n=4 $
An $\alpha$ -particle of energy ${1\over2} mv^2$ bombards by a heavy nuclear target ofcharge ze.Then the distance of closet approach for the alpha nucleus will be Proportional to
at distance of closest approach K.E = P.E ${1\over 2}mv^2 = {1 \over 4 \pi \epsilon_o} {(ze) (ze) \over ro} \Rightarrow ro = {ze^2 \over \pi epsilon_o mv^2 }$
An $\alpha $ -particle of energy 5 MeV is scattered though 180 by a fixed uranium nucleus. The distance of the closest approach nucleus The distance of the closest approach is of the order of
The distance of closest approach in a Rutherford scattering experiment can be estimated using the formula:
$$ d = rac{1}{4 \\pi \\epsilon_0} \\frac{2Ze^2}{E} $$
where $Z$ is the atomic number of the uranium nucleus (92), $e$ is the elementary charge, and $E$ is the energy of the alpha particle (5 MeV). Substituting these values, the order of magnitude for the distance of closest approach comes out to be approximately $10^{-12}$ cm.
which of the following transition in hydrogen atoms emits Photon of highest frequency?
n = 2 to n = 6 absorbs photon n = 9 to n = 2 absorbs photon n = 6 to n = 2 emission of photon n = 2 to n = 1 emission of photon $E = E_2-E_1 = {-13.6 \over 36}-({-13.6 \over 4 }) = -0.38+13.6 = 3.02 eV $ $ \triangle \Sigma^1 = E_2-E_1 = {-13.6 \over 4}-({-13.6 \over 1}) = -3.4+13.6 = 10.2 eV $ $ \triangle \Sigma^1 > \triangle E$ $hf' > hf $
An electron Passing through a Potential difference of 4.9 v collides with a mercury atom and transfer it to the first excited state what is transfer it to the first excited state. what is the wave length of Photon corresponding to the franition of mercury atom to its normal state.
energy of electron = ve $ E = 4.9 \times 1.6 \times 10 ^ {-19} J $ ${ hc \over \lambda} \Rightarrow \lambda = {hc \over E} $
Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is $1.67 \times 10^{-27}$ kg. (i) The energy of Photon causing the Photo electric emission is
$ K_{max} = hf - \phi $ $ hf = K_{max} + \phi $
Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is $1.67 \times 10^{-27}$ kg. (ii) The quantum number of the two leVels in the emission of the Photons are
Correspnding energy level for 2.55 eV is n = 2 , n = 4
Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is $1.67 \times 10^{-27}$ kg. (iii) In this transition change in the angular momentum of electron is ( where h is Plank Constant )
$Change \,Angular \, momentum = {4h \over 2 \pi } - {2h \over 2 \pi} \Rightarrow {2h \over \pi} - { h \over \pi } \Rightarrow {h \over \pi }$
Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is $1.67 \times 10^{-27}$ kg. (iv) The recoil speed of emitting atom causing that is atleast before the transition is of order of
According to conservation of momentum momentum of photon = momentum of lecoil altom ${ h \over \lambda } = mu $ $ \upsilon = { h \over m \lambda } = { hf \over mc } = E/mc $
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