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Right option is (b) 4 TIMES

The best EXPLANATION: For the FIRST excited level, n = 2.

r2 = (2)^2r0 = 4r0.

So, when a hydrogen atom is in its first excited level, its radius is 4 times of the Bohr radius.

The correct CHOICE is (c) 10^-11 m

To explain: The radius of an electron orbit in a hydrogen atom is of the order of 10^-11 m. It is equal to the most PROBABLE DISTANCE between the NUCLEUS and the electron in a hydrogen atom in its ground STATE.

The correct option is (d) When the electron is at its ground state

Explanation: The hydrogen atom is stable when the electron rests at the ground level of ENERGY, i.e. when the principal quantum number, n = 1. In other WORDS, when the electron is revolving in the FIRST orbit around the NUCLEUS, the hydrogen atom is stable.

Correct choice is (a) TRUE

The EXPLANATION is: YES, this is a true statement. Bohr’s model works only for elements like hydrogen because the model only considers the interactions between one electron and the nucleus. If there are more electrons, then they will REPEL the one electron, and this, in TURN, will change its energy level.

The CORRECT option is (c) de – Broglie SUGGESTED that electrons behaved like a wave

The best explanation: de – Broglie hypothesis did MODIFY Bohr’s second postulate. This postulate of Bohr regarding the QUANTIZATION of the angular momentum of an electron was further explained by Louis de Broglie. ACCORDING to de – Broglie, a moving electron in its circular orbit behaves like a particle-wave.

Right OPTION is (a) 5:9

To explain I would SAY: For a wavelength of BALMER series,

\(\frac {1}{\lambda }\) = R\( [ \frac {1}{4} – \frac {1}{9} ] \)

\(\frac {1}{\lambda } = \frac {5R}{30}\).

\(\frac {\lambda_{min}}{\lambda_{MAX}} = \frac {5R}{36} \times \frac {4}{R} \)

\(\frac {\lambda_{min}}{\lambda_{max}} = \frac {5}{9}\)

The correct CHOICE is (c) 1s^22s^22p^63s^23p^2

To EXPLAIN I would SAY: The atomic NUMBER of silicon is 14.

Therefore, ^14Si will have the following ELECTRONIC configuration:

1s^22s^22p^63s^23p^2

Right answer is (a) Assumes that the angular MOMENTUM of electrons is quantized

For explanation: BOHR MODEL of atoms assumes that the angular momentum of electrons is quantized. The atom is held between the nucleus and surroundings by electrostatic forces. The other options are not VALID.

The CORRECT choice is (a) 1.5 eV

Easy explanation: From THIRD EXCITED STATE, E = \(\frac {-13.6}{16}\)

E = -0.85 eV

Energy required to ionize H-atom from SECOND excited state = 0 – (-0.85)

E = +0.85 eV

Correct OPTION is (a) L = \(\frac {nh}{2p}\)

To elaborate: According to BOHR’s SECOND postulate, an electron revolves around the nucleus of an atom in orbits and, therefore, the angular momentum of the revolution is an integral multiple of \(\frac {h}{2p}\), where h is the Planck’s constant. Thus, the EXPRESSION for angular momentum is given as:

L = \(\frac {nh}{2p}\)

Right option is (b) String

To elaborate: A string is USED to represent the BEHAVIOR of a particle analogously to the waves traveling on it.Particle waves can lead to standing waves held under resonant CONDITIONS. When a stationary string is PLUCKED, it causes several wavelengths to be excited. But, we know that only the ones which have nodes at the ends will survive.

The correct option is (B) BOHR theory

The best I can explain: Niels Bohr explained the line spectrum of the hydrogen atom with the assumption that electrons revolved around an atom in CIRCULAR orbits and that the orbit closer to the nucleus represented the ground state and the FARTHER orbits represented the HIGHER levels of energy.

The correct option is (d) An electron will ABSORB ENERGY when JUMPING from the 1^st ORBIT to the 3^rd orbit

To explain I would say: An electron will absorb energy when jumping from the 1^st orbit to the 3^rd orbit. Only by absorbing energy, an electron will be able to jump from the first orbit to the THIRD orbit in the atomic spectrum.

The correct choice is (a) True

The best I can explain: Yes, this is a true statement. In a string, STANDING waves are formed only when the total DISTANCE TRAVELED by a WAVE is an integral number of wavelengths. In this way, distance and wavelength are PROPORTIONAL to each other. Hence, the expression is given as:

2πrk = kλ

The correct choice is (a) Energy

To explain I would say: According to the UNCERTAINTY principle,

ΔE.Δt >= \(\frac {h}{2\pi }\).

THUS the time MEASURED will BECOME uncertain if ΔE is measured with high certainty.

Right answer is (B) False

Best explanation: The EQUATION is GIVEN as:

rn = R1 n^2

r2 = (2)^2r0 = 4r0. Thus, in terms of BOHR radius a0, the radius of the second Bohr orbit of a hydrogen atom is given by 4 a0.

Right choice is (c) s-ELECTRON

The best explanation: The valence electron in an alkali metal is an s-electron. GENERALLY, they make up Group 1 of the periodic table. The DIFFERENT examples that come under this CATEGORY are lithium, potassium, and FRANCIUM.

Correct choice is (c) Multiple

For explanation I WOULD say: Even though hydrogen has only one electron in its OUTERMOST shell, it has multiple spectral LINES. This is because hydrogen has many energy LEVELS. When the electron excites from a lower level of energy into a higher level, then it RELEASES a photon, and this photon is the one that appears as spectral lines.

The CORRECT CHOICE is (a) \(\frac {9}{16}\)

The explanation: \(\frac {KE_{LI}}{KE_{Be}} = \big [ \frac {(\frac {Z_{Li}}{2} )}{(\frac {Z_{Be}}{2})} \big ]^2 \)

\(\frac {KE_{Li}}{KE_{Be}} = \big [ \frac {(\frac {3}{2} )}{(\frac {4}{2})} \big ]^2 \)

\(\frac {KE_{Li}}{KE_{Be}} = \frac {9}{16}\).

The correct answer is (d) 2

The EXPLANATION: N = 4.

The number of spectral lines emitted = \(\frac {n(n-1)}{2}\).

\(\frac {n (n-1)}{2} = \frac {(4 \times 3)}{6}\) = 2

Right choice is (a) The transition of ELECTRONS between two ENERGY levels

To elaborate: The observed spectral lines are CAUSED by the transition of electrons between two energy levels in an atom. The emission spectrum of the HYDROGEN atom is DIVIDED into many spectral series, with wavelengths that are given by Rydberg’s formula.

The correct ANSWER is (b) False

To elaborate: Bohr model assumes that the angular momentum of electrons is quantized. THEREFORE, the Bohr model of the atoms involved is independent of EINSTEIN’s photoelectric equation.

Correct ANSWER is (B) 10^-12cm

To ELABORATE: r0 = \(\frac {(2Ze^2)}{(4\pi \varepsilon_0 (\frac {1}{2}m v^2))}\)

r0 = \(\frac {9 \times 10^9 \times 2 \times 92 \times (1.6 \times 10^{-19})^2}{10 \times 1 \times 10^{-13}}\) J

r0 = 4.239 × 10^-14 m

r0 = 4.2 × 10^-12 CM.

The correct choice is (d) de – Broglie concluded that wavelengths of matter WAVES can be quantized

Explanation: de – Broglie concluded his modification of Bohr’s SECOND POSTULATE by stating that the wavelengths of matter waves can be quantized. This implies that the ELECTRONS can exist in those orbits which had a complete set of SEVERAL wavelengths.

The correct choice is (c) Lyman series

The EXPLANATION: The TRANSITIONS that end at the ground LEVEL, where the principal QUANTUM number is one, are called the Lyman series. Here, the energies that are released are so large, and therefore, these spectral lines appear in the ultraviolet region of the spectrum.

Correct choice is (a) Doubly ionized LITHIUM

Best EXPLANATION: The EQUATION is given as:

r = \(\frac {h^2}{2\pi^2 m k Z e^2}\)

Hence, of the given atoms/ions, (Z = 3) is maximum for doubly ionized lithium, so the radius of its first orbit is MINIMUM.

The CORRECT answer is (c) \(\frac {1}{n^2}\)

EXPLANATION: The EQUATION is GIVEN as:

rn = \(\frac {n^2h^2}{4\pi^2mkZe^2}\).

From this, we can understand that rn is directly proportional to n^2.

Correct CHOICE is (a) Zero

Easy explanation: The perpendicular distance between the path of a projectile and the CENTER of the potential field is the impact parameter. Therefore, for a head-on collision of the α-particle with a nucleus, the impact parameter is EQUAL to zero.

Right option is (d) C and N^+

To EXPLAIN I WOULD say: Carbon and the positive ion of NITROGEN (N^+) will have the same ELECTRONIC configuration.

The electronic configuration of both Carbon and the positive ion of nitrogen is as follows:

 1s^22s^22p^6.

The CORRECT OPTION is (d) r proportional to n^2

The best explanation: The equation is given as:

r = \(\frac {n^2 h^2}{4\pi^2 m K Z e^2}\)

Therefore, we can say that the radius of the ORBIT is directly proportional to the square of the PRINCIPAL quantum number.