(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.For the ground state,$n = 1$,so $E_1 = -13.6 \ eV$.Excited states correspond to $n > 1$ (i.e.,$n = 2, 3, 4, \dots$).For the first excited state,$n = 2$:$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.Thus,$-3.4 \ eV$ is a possible energy value for an excited state.

Class 11 Chemistry Structure of Atom Atomic models and Planck's quantum theory

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The energy of an electron in the first Bohr orbit of $H$ atom is $-13.6 \ eV$. The possible energy value$(s)$ of the excited state$(s)$ for electrons in Bohr orbits of the hydrogen atom is(are) $............ \ eV$.

IIT 1998 All IIT

A
$-3.4$
B
$-4.2$
C
$-6.8$
D
$+6.8$

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Structure of Atom

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The energy of an electron in the first Bohr orbit of $H$ atom is $-13.6 \ eV$. The possible energy value$(s)$ of the excited state$(s)$ for electrons in Bohr orbits of the hydrogen atom is(are) $............ \ eV$.

Similar Questions

$A$ $124 \ W$ bulb converts only $15 \ \%$ of the energy supplied to it into visible light of wavelength $640 \ \text{nm}$. How many photons are emitted by the light bulb in one second?

The energy of a photon is $3 \times 10^{-12} \ erg$. What is its wavelength in $nm$? $(h = 6.62 \times 10^{-27} \ erg \cdot s; \ c = 3 \times 10^{10} \ cm/s)$

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