NEET Atomic Structure MCQs & PYQs

Practice NEET Atomic Structure Questions

Question 61:

easy

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is:

(2011 Pre)

For first line of Lyman for H-atom ($n=2 \rightarrow 1$), $1/\lambda = R(1/1^2 - 1/2^2) = 3R/4$. For second line of Balmer for ion ($n=4 \rightarrow 2$), $1/\lambda = Z^2 R(1/2^2 - 1/4^2) = Z^2 R(3/16)$. Equating them gives $3R/4 = Z^2 R(3/16) \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

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Question 62:

easy

The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping potential is estimated to be: (the energy of the electron in $n^{th}$ state)

(2010 Mains)

Energy of emitted photon $E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 eV$. Using Einstein's photoelectric equation, max kinetic energy $K_{max} = E - \phi = 12.09 - 5.1 = 6.99 eV \approx 7 eV$. Hence, the stopping potential is $7 V$.

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Question 63:

easy

The energy of a hydrogen atom in the ground state is -13.6 eV. The energy of a $He^+$ ion in the first excited state will be:

(2010 Pre)

The energy of an electron in a hydrogen-like ion is $E_n = -13.6 \frac{Z^2}{n^2} eV$. For a $He^+$ ion, $Z=2$. The first excited state corresponds to $n=2$. Thus, $E_2 = -13.6 \frac{2^2}{2^2} = -13.6 eV$.

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Question 64:

easy

24. The ionisation energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between: (2009)

Number of spectral lines emitted is $\frac{n(n-1)}{2} = 6 \Rightarrow n = 4$. Maximum wavelength corresponds to the minimum energy difference. For transitions among levels up to $n=4$, the transition $4 \rightarrow 3$ has the minimum energy difference and thus the maximum wavelength.

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Question 65:

easy

The ground state energy of hydrogen atom is -13.6 eV. When its electron is in the first excited state, its excitation energy is:

(2008)

Excitation energy is the energy required to excite the electron from the ground state ($n=1$) to a particular state. The energy of the first excited state ($n=2$) is $-13.6/4 = -3.4 eV$. The excitation energy is $-3.4 - (-13.6) = 10.2 eV$.

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Question 66:

easy

The total energy of electron in the ground state of hydrogen atom is -13.6 eV. The kinetic energy of an electron in the first excited state is:

(2007)

The total energy in the first excited state ($n=2$) is $E_2 = -13.6/2^2 = -3.4 eV$. Since kinetic energy $K = -E_n$, the kinetic energy in the first excited state is $3.4 eV$.

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Question 67:

easy

Hydrogen atoms are excited from ground state of the principle quantum number 4. Then the number of spectral lines observed will be:

(1993)

The number of possible spectral lines emitted when transitioning from the nth state to the ground state is $\frac{n(n-1)}{2}$. For $n=4$, the number of lines is $\frac{4 \times 3}{2} = 6$.

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Question 68:

easy

Which source is associated with a line emission spectrum?

(1993)

Line emission spectra are characteristic of excited atoms in low-pressure gases. A neon street sign contains low-pressure neon gas which, when excited, emits a characteristic line spectrum.

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Question 69:

easy

In terms of Bohr radius $a_0$, the radius of the second Bohr orbit of a hydrogen atom is given by:

(1992)

The radius of the nth Bohr orbit for a hydrogen atom is $r_n = a_0 n^2$. For the second orbit ($n=2$), the radius is $r_2 = a_0 (2^2) = 4a_0$.

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Question 70:

easy

Energy E of a hydrogen atom with principal quantum number n is given by $E = \frac{-13.6}{n^2} eV$. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately:

(2004)

The energy of the emitted photon is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \left(\frac{1}{4} - \frac{1}{9}\right) eV$. This gives $\Delta E = 13.6 \times \frac{5}{36} = 1.88 eV \approx 1.9 eV$.

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