Atoms Test

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Atoms Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

In which of the following systems will the radius of the first orbit of the electron be smallest?

A
hydrogen

B
singly ionized helium

C
deuteron

D
tritium

Solution

Hydrogen
Question 2
1 / -0

In which of these, will radius of first orbit be minimum?

A
$$H$$ atom

B
$$^{2}_{1}H$$ atom

C
$$He^{+}$$ ion

D
$$Li^{2+}$$ ion

Solution

$$r_{n}^{1}=4\pi \epsilon _{0}\dfrac{n^{2}h^{2}}{4\pi ^{2}z.me^{2}}$$
$$Z$$ is max for $$Li^{2+}$$ ion.
$$\therefore$$ radius is minimum for $$Li^{2+}$$ ion.
Question 3
1 / -0

In which part of the electromagnetic spectrum the Lyman series of hydrogen is found ?

A
ultraviolet

B
infrared

C
visible

D
X-ray region

Solution
Question 4
1 / -0

Directions For Questions

The basic idea of Quantum Mechanics is that motion in any system is quantized.The system obeys Classical Mechanics except that not every motion which obeys the Bohr-Sommerfeld Quantization,$$\oint \vec{P}.\vec{\mathrm{d} r}=nh, n \in N,$$ where $$\vec{P}$$ is the momentum,$$\vec{r}$$ is the position vector and the integral is carried over a closed path.Assuming this is applicable to a particle of mass $$m$$ moving with a constant speed in a box of length $$L$$ having elastic collisions with the walls of the box, answer the following questions:
($$h=6.6\times 10^{-34} J-sec,L=3.3 A^{\circ},m=10^{-30}kg$$)

...view full instructions

The allowed momenta are given by

A
$$\dfrac{nh}{2L}$$

B
$$\dfrac{nh}{L}$$

C
$$\dfrac{nh}{2\Pi L}$$

D
$$\dfrac{nh}{4L}$$

Solution

From Bohr-Sommerfeld Quantization condition given in the paragraph, for a back and forth motion.
$$P\times 2L=nh$$, where P is the momentum,
$$P=\dfrac{nh}{2L}$$
Question 5
1 / -0

Directions For Questions

The basic idea of Quantum Mechanics is that motion in any system is quantized.The system obeys Classical Mechanics except that not every motion which obeys the Bohr-Sommerfeld Quantization,$$\oint \vec{P}.\vec{\mathrm{d} r}=nh, n \in N,$$ where $$\vec{P}$$ is the momentum,$$\vec{r}$$ is the position vector and the integral is carried over a closed path.Assuming this is applicable to a particle of mass $$m$$ moving with a constant speed in a box of length $$L$$ having elastic collisions with the walls of the box, answer the following questions:
($$h=6.6\times 10^{-34} J-sec,L=3.3 A^{\circ},m=10^{-30}kg$$)

...view full instructions

The allowed kinetic energy of the particle is

A
$$\dfrac{n^{2}h^{2}}{8\pi ^{2}mL^{2}}$$

B
$$\dfrac{n^{2}h^{2}}{2mL^{2}}$$

C
$$\dfrac{n^{2}h^{2}}{8mL^{2}}$$

D
$$\dfrac{n^{2}h^{2}}{32mL^{2}}$$

Solution

From Bohr-Sommerfeld Quantization condition given in the paragraph, for a back and forth motion.
$$P\times 2L=nh$$, where P is the momentum,
$$P=\dfrac{nh}{2L}$$

And the kinetic energy, $$ E=\dfrac{P^{2}}{2m}$$
$$E=\dfrac{n^{2}h^{2}}{8mL^{2}}$$
Question 6
1 / -0

Assume an imaginary world where angular momentum is quantized to even multiple. The longest possible wave length emitted by hydrogen in the visible spectrum is

A
$$484 nm$$

B
$$300 nm$$

C
$$400 nm$$

D
$$350 nm$$

Solution

The angular momentum is quantized to even multiple of $$\bar h$$, hence
$$mvr = 2n \hbar $$
$$mv = \dfrac{2n\hbar }{r}$$
$$mv^2 = \dfrac{m^2v^2}{m} = \dfrac{(2n\hbar )^2}{mr}$$

The Coulomb's force of attraction is equal to the centripetal force.
$$\dfrac{Ze^2}{4\pi \epsilon_0 r^2} = \dfrac{mv^2}{r}$$
$$\dfrac{Ze^2}{4\pi \epsilon_0 r^2} = \dfrac{\dfrac{(2n\hbar )^2}{mr}}{r}$$
$$r = \dfrac{(2n\hbar )^2 4\pi \epsilon_0}{mZe^2}$$

Now, the binding energy of electron is
$$BE = \dfrac{-Ze^2}{8\pi \epsilon_0 r}$$
$$BE = \dfrac{-Ze^2}{8\pi \epsilon_0 (\dfrac{(2n\hbar )^2 4\pi \epsilon_0}{mZe^2})}$$
$$BE = \dfrac{-Z^2e^4m}{32 \pi \epsilon_0^2n^2h^2}$$

$$BE = \dfrac{-3.4}{n^2} eV$$

For longest possible wavelength in visible spectrum $$n_2 = 4$$ and $$n_1 = 1$$
$$h\nu = -3.4[1 - \dfrac{1}{4}]$$
$$h\nu = -3.4 (\dfrac{3}{4})$$
$$\nu = -\dfrac{3.4 \times 0.75}{h}$$

Therefore, the longest possible wavelength emitted by hydrogen in the visible spectrum is(in nm)
$$\lambda = \dfrac{c}{\nu}$$
$$\lambda = \dfrac{hc}{3.4 \times 0.75}$$

$$\lambda = \dfrac{1236}{2.55}$$

$$\lambda = 484\ nm$$
Question 7
1 / -0

A particle of mass $$m$$ moves around in a circular orbit in a centro symmetric potential field u(r)$$=\dfrac{kr^{2}}{2}$$. Using Bohr’s quantization rule, the permissible energy levels are

A
$$\sqrt{\dfrac{k}{m}}$$

B
$$n\sqrt{\dfrac{k}{m}}$$

C
$$n$$

D
$$\sqrt{\dfrac{nk}{m}}$$

Solution

$$U=\dfrac{kr^2}{2}$$
$$\implies F=-\dfrac{mv^2}{r}=\dfrac{-dU}{dr}=-kr$$
$$\implies \dfrac{1}{2}mv^2=\dfrac{1}{2}kr^2$$
Total energy $$=\dfrac{1}{2}mv^2+U=kr^2$$

From bohr's quantization, $$mvr=\dfrac{nh}{2\pi}$$
$$\implies m^2v^2r^2=\dfrac{n^2h^2}{4\pi^2}$$
$$\implies mkr^4=\dfrac{n^2h^2}{4\pi^2}$$
$$\implies r^2=\dfrac{nh}{2\pi \sqrt{mk}}$$
Energy $$=kr^2=k\dfrac{nh}{2\pi \sqrt{mk}}\propto n\sqrt{\dfrac{k}{m}}$$
Question 8
1 / -0

Let $$\mathrm{v}_{1}$$ be the frequency of the series limit for Lyman series, $$\mathrm{v}_{2}$$ the frequency of the first line of the Lyman series and $$\mathrm{v}_{3}$$ the frequency of the series limit of Balmer series. Then which of the following relations is true?

A
$$\mathrm{v}_{1}-\mathrm{v}_{2}=\mathrm{v}_{3}$$

B
$$\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{3}$$

C
$$\displaystyle \mathrm{v}_{3}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}$$

D
$$\mathrm{v}_{1}+\mathrm{v}_{2}=\mathrm{v}_{3}$$

Solution

Series limit means the shortest possible wavelength (maximum photon energy) and first line means the longest possible wavelength, (minimum photon energy) in the series

$$\displaystyle \mathrm{v}=\mathrm{c}[\frac{1}{\mathrm{n}^{2}}-\frac{1}{\mathrm{m}^{2}}]$$, where c is a constant.

For series limit of Lyman series $$\mathrm{n}=1 \mathrm{m}=\infty \mathrm{v}_{1}=\mathrm{c}$$

For first line of Lyman series $$\mathrm{n}=1 \mathrm{m}=2 \displaystyle \mathrm{v}_{2}=\frac{3\mathrm{c}}{4}$$

For series limit of Balmer serles $$\mathrm{n}=2 \mathrm{m}=\infty \displaystyle \mathrm{v}_{3}=\frac{\mathrm{c}}{4}$$
:. $$\mathrm{v}_{1}-\mathrm{v}_{2}=\mathrm{v}_{3}$$

$$E_1 = E_2 + E_3$$
$$hV_1 = hV_2 + hV_3$$
$$V_1 = V_2 + V_3$$
Question 9
1 / -0

The radius of hydrogen atom is 0.53 $$\mathop A\limits^0$$. The radius of $$Li^{2+}$$ is of

A
$$1.27\mathop A\limits^0 $$

B
$$0.17 \mathop A\limits^0$$

C
$$0.57 \mathop A\limits^0$$

D
$$0.99 \mathop A\limits^0$$

Solution

$$\begin{array}{l}\text { radius, } r=\frac{n^{2} h^{2}}{4 \pi^{2} m z e^{2}}=0.53\frac{n^{2}}{z} \dot A \\\text { for ground state } n=1 . \\\text { radius hydrogen in ground state } \\\text { = }\frac{0.53(1)^{2}}{1}=0.53 \dot A \\\text { [atomic no. of } \\\text { hydrogen, } z=1]\end{array}$$
$$\begin{aligned}\text { that of } & L i^{2+}\text { is } \\r=& \frac{0.53 \times(1)^{2}}{3} \\&=0.177 \mathrm{~A} \\& \approx 0.17\mathrm{~A} .\end{aligned}$$
Question 10
1 / -0

Directions For Questions

The basic idea of Quantum Mechanics is that motion in any system is quantized.The system obeys Classical Mechanics except that not every motion which obeys the Bohr-Sommerfeld Quantization,$$\oint \vec{P}.\vec{\mathrm{d} r}=nh, n \in N,$$ where $$\vec{P}$$ is the momentum,$$\vec{r}$$ is the position vector and the integral is carried over a closed path.Assuming this is applicable to a particle of mass $$m$$ moving with a constant speed in a box of length $$L$$ having elastic collisions with the walls of the box, answer the following questions:
($$h=6.6\times 10^{-34} J-sec,L=3.3 A^{\circ},m=10^{-30}kg$$)

...view full instructions

The difference between I and II energy levels is

A
$$ 1.5\times 10^{-18} J$$

B
$$ 3\times 10^{-18} J$$

C
$$ 6\times 10^{-18} J$$

D
$$ 12\times 10^{-18} J$$

Solution

For this particle in the box: $$E=\dfrac{n^{2}h^{2}}{8mL^{2}}$$

The difference between I and II energy levels is:
$$E_{diff}=\dfrac{3h^{2}}{8mL^{2}}=1.5 \times 10^{-18} J$$

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Atoms Test - 66

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Atoms Test - 66

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SHARING IS CARING

Question 1

hydrogen

singly ionized helium

deuteron

tritium

Solution

Question 2

$$H$$ atom

$$^{2}_{1}H$$ atom

$$He^{+}$$ ion

$$Li^{2+}$$ ion

Solution

Question 3

ultraviolet

infrared

visible

X-ray region

Solution

Question 4

$$\dfrac{nh}{2L}$$

$$\dfrac{nh}{L}$$

$$\dfrac{nh}{2\Pi L}$$

$$\dfrac{nh}{4L}$$

Solution

Question 5

$$\dfrac{n^{2}h^{2}}{8\pi ^{2}mL^{2}}$$

$$\dfrac{n^{2}h^{2}}{2mL^{2}}$$

$$\dfrac{n^{2}h^{2}}{8mL^{2}}$$

$$\dfrac{n^{2}h^{2}}{32mL^{2}}$$

Solution

Question 6

$$484 nm$$

$$300 nm$$

$$400 nm$$

$$350 nm$$

Solution

Question 7

$$\sqrt{\dfrac{k}{m}}$$

$$n\sqrt{\dfrac{k}{m}}$$

$$n$$

$$\sqrt{\dfrac{nk}{m}}$$

Solution

Question 8

$$\mathrm{v}_{1}-\mathrm{v}_{2}=\mathrm{v}_{3}$$

$$\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{3}$$

$$\displaystyle \mathrm{v}_{3}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}$$

$$\mathrm{v}_{1}+\mathrm{v}_{2}=\mathrm{v}_{3}$$

Solution

Question 9

$$1.27\mathop A\limits^0 $$

$$0.17 \mathop A\limits^0$$

$$0.57 \mathop A\limits^0$$

$$0.99 \mathop A\limits^0$$

Solution

Question 10

$$ 1.5\times 10^{-18} J$$

$$ 3\times 10^{-18} J$$

$$ 6\times 10^{-18} J$$

$$ 12\times 10^{-18} J$$

Solution

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