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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Atoms

In a transition to a state of excitation energy 10.19eV , a hydrogen atom emits a $\;4890 A^{0}\;$ photon . Then the binding energy of initial state is :

$(a)\;-1.2 eV\qquad(b)\;-2.4 eV\qquad(c)\;-0.87 eV\qquad(d)\;-3.4 eV$

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1 Answer

Answer : (c) -0.87 eV
Explanation :
The energy of emitted photon is
$h \nu =\large\frac{hC}{\lambda}=\large\frac{12.4\times10^{3}eV\;A^{0}}{4.89\times10^{3}A^{0}}$
$=2.54eV$
The excitation energy $\;(\varepsilon_{x})\;$ is energy to excite the atom to a level above the ground state Therefore , The energy of level is
$\varepsilon_{n}=\varepsilon_{1}+\varepsilon_{x}=-13.6eV+10.19eV$
$=-3.41eV$
The photon arises from transition between energy status such that $\;\varepsilon_{x}-\varepsilon_{1}=h \nu\;:hence$
$\varepsilon_{x}-(-3.41eV)=2.54eV$
$or\quad \; \varepsilon_{x}=-0.87eV\;.$
answered Feb 22, 2014 by yamini.v
 

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