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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.

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Explanation

energy of electron = ve $ E = 4.9 \times 1.6 \times 10 ^ {-19} J $ ${ hc \over \lambda} \Rightarrow \lambda = {hc \over E} $

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The speed of daughter nuclei is

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Explanation

$ {1 \over 2} Mv^2 = ( \triangle m)C^2$ $ V^2 = {2(\triangle m) C^2 \over M}$ $V = C {\sqrt{2(\triangle m)} \over m}$

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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

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Explanation

$ K_{max} = hf - \phi $ $ hf = K_{max} + \phi $

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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

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Explanation

Correspnding energy level for 2.55 eV is n = 2 , n = 4

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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 )

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Explanation

$Change \,Angular \, momentum = {4h \over 2 \pi } - {2h \over 2 \pi} \Rightarrow {2h \over \pi} - { h \over \pi } \Rightarrow {h \over \pi }$

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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

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Explanation

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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A and B are two radioactive substance whose half lives are 1 and 2 years respectively. Initially 10 g of A and 1 g of B is taken. The time after which they will have same quantity remaining is

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Explanation

$ {m_1 \over mb_1} = ( {1\over 2}) ^{1\over T1/2} = ({1 \over 2}) ^{t \over 1} = { 1 \over 2^t}$ $ m_1 = {10 \over 2^+} $ $ m_1 = m_2 $ $ m_2 = {1 \over 2^{t /2}}$ ${10 \over 2^t } = {1 \over 2^{t/2}}$ $ 10 = (2)^{t/2}$ $ log 10 = t/2 log 2 $ $1.0000 = {t \over 2} \times 0.3010 $ $ t =6.64 year \approx 6.6 year $

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which of these is a fusion reaction

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Explanation

$_1^3H +_1^2H = _2^4 He + \,_0n^1$

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The activity of a sample of a radio- active material is at time $t_1$ and $ A_2$ at time $ t_2$ (where $ t_2 > t _1 $) if T its mean life is then

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Explanation

$A_1 = A_0 e^{-t_1/t}$ $A=A_0^ {e -\lambda t } = A = A_0e^{-t/t}$ $A_0 = {A_1 \over e-t_1/T} \Rightarrow A_0 = {A_1 e^{t_1 /T}} $ $ Now , A_2 = A_0e^{-t_2/T}$ $ =A_1 e^{t_1/T} e^{-t_2/T}$ $ A_2 = A_1 e(t_1 - t_2)/ T$

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In the following dis-integlation series $ _{92} U ^ {238} \rightarrow x \rightarrow _z y^A$ The value of Z and A respectively will be

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Explanation

$_{92}U^{238} \rightarrow x \rightarrow _zY^A$ $_{92}U^{238} \rightarrow _{90}Th^{234} \rightarrow _{91}Y^{234}$

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