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

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 $\;\varepsilon_{x}-\varepsilon_{1}=h \nu\;:hence$

$\varepsilon_{x}-(-3.41eV)=2.54eV$

$or\quad \; \varepsilon_{x}=-0.87eV\;.$

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