An electron collides with a dfixed hydrogen atom in its ground stat. Hydrogen atom gets excited and the collding electron loses all its kinetic energy. Consequently the hydrgogen atom may emits a photon corresponding to the alrgest wavelength of the Balmer series. The K.E. of colliding electron will be 24.2/N eV. Find the value of N.
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An electron of a hydrogen like atom is in excited, state. If total energy of the electron is -4.6 eV, then evaluate (i) the kinetic energy and (ii) th
Given ,total energy of the electron `E=-4.6 EV` (i) Kinetic energy of electron K=-(Total energy E,) `rArr " " =-E=-(-4.6)=4.6 eV` (ii) de- Brogile wavelength `lambda_(d)=(h)/(sqrt(2 mK))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx4.6xx1.6xx10^(-19)))` `0.57xx10^(-9)m=0.57 nm`
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The energy of the electron in the ground state of hydrogen atom is `-13.6 eV`. Find the kinetic energy and potential energy of electron in this state.
Correct Answer - B Kinetic energy of electron in ground state =-(Totoal energy in this state) `" " --(13.6 )eV=13.6 eV`
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In an inelastic collision an electron excites a hydrogen atom from its ground state to a M-Shell state. A second electron collides instantaneously wit
Correct Answer - D The ground state of hydrogen (n=1) is represented by K, the first excited state (n=2) is represented by L,the second excited state (n=3) is represent byb...
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An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the r
Correct Answer - B `E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For second excited state to first excited state `E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))` For first excited state...
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An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energ
Correct Answer - C Excitation upto `n=3` is required so that visible length is emitted upon de-excitation. So required energy `=13.6 (1-1/9)=12.1 eV`
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Calculate the energy required to excited one litre of hydrogen gas at `1` atm and `298 K` to the excited state of atomic hydrogen. The energy for the
Correct Answer - B Mole of `H_(2)` present in one litre `=(PV)/(RT)=(1xx1)/(0.0821xx298)=0.0409` Thus, energy needed to break H-H bonds in `0.0409` mole `H_(2)=0.0409xx436=17.83 kJ`. Also energy needed to excite one H...
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`H^(+)` ions are in excited state from where maximum 6 types of photons can be emitted. If E is the separation energy of electron in this excited stat
Correct Answer - 4 `6=(n(n-1))/(2)" "n=4` Separation energy, `E=+13.6 xx(z^(2))/(n^(2))=+13.6xx((2)^(2))/((4)^(2))=3.4 eV` l.E. of Hydrogen atom `=x xxE=13.6 eV` `x=(13.6)/(E)=4`
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A hydrogen like atom (atomic number = Z) is in higher excited state of quantum number n. This excited atom can make a transition to first excited stat
Correct Answer - `n = 7 ; Z = 3`
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A free hydrogen atom in its ground state is at rest. A neutron having kinetic energy `k_(0)` collides head one with the atom. Assume that mass of both
Correct Answer - (a) `20.4 eV` , (b) `8.72 eV` or `4.19 eV`
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A hydrogen atom in ground state is moving with a kinetic energy of 30 eV. It collides with a deuterium atom in ground state at rest. The hydrogen atom
Correct Answer - `20 eV, 13.2 eV`
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