Physics Test

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Physics Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

A hydrogen atom and $L i^{+ +}$ ion are both in the second excited state. If $l_{H}$ and $l_{L i}$ are their respective electronic moments, and $E_{H}$ and $E_{L i}$ their respective energies, then:

A
$ℓ_{H} > ℓ_{L i}$ and $| E_{H} | > | E_{L i} |$

B
$ℓ_{H} = ℓ_{L i}$ and $| E_{H} | < | E_{L i} |$

C
$ℓ_{H} = ℓ_{L i}$ and $| E_{H} | > | E_{L i} |$

D
$ℓ_{H} < ℓ_{L i}$ and $| E_{H} | < | E_{L i} |$

Solution

Given, $Z_{H} = 1, Z_{L i} = 3$ and $n_{H} = n_{L i} = 3$
1. Angular momentum is independent of $Z$ , therefore $l_{H} = l_{L i} $
2. Energy is directly proportional to the square of atomic number, hence $E_{H} < E_{L i }$
Question 2
1 / -0

Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength $λ$ (given in terms of Rydberg constant $R$ for the hydrogen atom) equal to:

A
$\frac{9}{5 R}$

B
$\frac{36}{5 R}$

C
$\frac{18}{5 R}$

D
$\frac{4}{R}$

Solution

Energy is related to mass: $E_{n} \propto m$
The largest wavelength $λ_{m a x}$ photon will correspond to the transition of the particle from $n = 3$ to the first excited state, $n = 2$ .
$\begin{array}{l} \frac{1}{λ_{\max}} = 2 R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) \\ \frac{1}{λ_{\max}} = \frac{5 R}{18} \\ ∴ λ_{\min} = \frac{18}{5 R} \end{array}$
Question 3
1 / -0

Hydrogen $(_{1} H^{1})$ , Deuterium $(_{1} H^{2})$ , singly ionized Helium ${(_{2} H e^{4})}^{+}$ and doubly ionized Lithium ${(_{3} L i^{6})}^{+ +}$ all have one electron around the nucleus. Consider an electron transition from $n = 2$ to $n = 1$ . If the wavelengths of emitted radiation are $λ_{1}, λ_{2}, λ_{3}$ , and $λ_{4}$ respectively, then approximately which of the following is correct?

A
$\lambda_{1}=\lambda_{2}=4 \lambda_{3}=9 \lambda_{4}$

B
$\lambda_{1}=2 \lambda_{2}=3 \lambda_{3}=4 \lambda_{4}$

C
$4 \lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}$

D
$\lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}$

Solution

From Rydberg's formula, $\frac{1}{λ} = R z^{2} (\frac{1}{1^{2}} - \frac{1}{2^{2}})$
$\begin{array}{l} ∴ λ = \frac{4}{3 R z^{2}} \\ λ_{1} = \frac{4}{3 R} \\ λ_{2} = \frac{4}{3 R} \\ λ_{3} = \frac{4}{12 R} \\ λ_{4} = \frac{4}{27 R} \\ \Rightarrow λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4} \end{array}$
Question 4
1 / -0

The intensity of X-ray from a Coolidge tube is plotted against wavelength as shown in the figure. The minimum wavelength found is ${\lambda_{c}}$ and the wavelength of $K_{\alpha}$ line is $\lambda_{k}$. As the accelerating voltage is increased:

A
$\lambda_{k}-\lambda_{c}$decreases

B
$\lambda_{k}-\lambda_{c}$increases

C
$\lambda_{k}$increases

D
$\lambda_{k}$decreases

Solution

Wavelength $λ_{k}$ is independent of the accelerating voltage $V$ , while the minimum wavelength $λ_{c}$ , is inversely proportional to $V$ .
Therefore as $V$ increases, $λ_{k}$ remains unchanged whereas $λ_{c}$ decreases of $λ_{k} - λ_{c}$ will increases.
Question 5
1 / -0

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the first members of Lyman and Paschen series, respectively, then $\lambda_{1}: \lambda_{2}$ is:

A
$1 : 3$

B
$1 : 30$

C
$7 : 50$

D
$7 : 108$

Solution

For the first-line of the Lyman series,
$\begin{array}{l} n_{1} = 1 and n_{2} = 3 \\ ∴ \frac{1}{λ_{1}} = R (\frac{1}{1^{2}} - \frac{1}{2^{2}}) \\ = R (1 - \frac{1}{4}) \end{array}$
$= \frac{3 R}{4}$ ...(1)
For first line of Paschen series
$\begin{array}{l} n_{1} = 3 and n_{2} = 4 \\ ∴ \frac{1}{λ_{2}} = R (\frac{1}{3^{2}} - \frac{1}{4^{2}}) \\ = R (\frac{1}{9} - \frac{1}{16}) \end{array}$
$= \frac{7 R}{144}$ ...(2)
From equation (1) and (2)
$∴ \frac{λ_{1}}{λ_{2}} = \frac{7 R}{144} \times \frac{4}{3 R}$
$= \frac{7}{108}$
Question 6
1 / -0

A diatomic molecule is made of two masses $m_{1}$ and $m_{2}$ , which are separated by a distance $r$ . If we calculate its rotational energy by applying Bohr rule of angular momentum quantization, then its energy will be given by:

A
$\frac{(m_{1} + m_{2}) 2 n^{2} h^{2}}{2 m_{1} 2 m_{2} 2 r^{2}}$

B
$\frac{n^{2} h^{2}}{2 (m_{1} + m_{2}) r^{2}}$

C
$\frac{2 n^{2} h^{2}}{(m_{1} + m_{2}) r^{2}}$

D
$\frac{(m_{1} + m_{2}) n^{2} h^{2}}{2 m_{1} m_{2} r^{2}}$

Solution

A diatomic molecule consists of two atoms of masses m₁ and m₂ at a distance r apart. Let r₁ and r₂ be the distances of the atoms from the center of mass.

The moment of inertia of this molecule about an axis passing through its center of mass and perpendicular to a line joining the atoms is
Question 7
1 / -0

Energy levels $A, B$ , and $C$ of a certain atom correspond to increasing values of energy $E_{A} < E_{B} < E_{C}$ . If $λ_{1}$ , $λ_{2}$ , and $λ_{3}$ are the wavelengths of radiations corresponding to transitions $C$ to $B, B$ to $A$ , and $C$ to $A$ , then which of the following relations is correct?

A
$λ_{1} + λ_{2} + λ_{3} = 0$

B
$λ_{3} = λ_{1} + λ_{2}$

C
$λ_{3} = \frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}}$

D
$λ_{3} = \frac{λ_{1} + λ_{2}}{2}$

Solution

As we know,
$E = h v = \frac{h c}{λ}$
$E_{3} - E_{1} = (E_{3} - E_{2}) + (E_{2} - E_{1})$
$\frac{h c}{λ_{3}} = \frac{h c}{λ_{1}} + \frac{h c}{λ_{2}}$
$\frac{1}{λ_{3}} = \frac{1}{λ_{1}} + \frac{1}{λ_{2}}$
$λ_{3} = \frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}}$
Question 8
1 / -0

The energy of an electron in the second orbit of a hydrogen atom in $E_{2}$ . The energy of an electron in the third orbit of $H e^{2}$ will be:

A
$\frac{9}{16} E_{2}$

B
$\frac{16}{9} E_{2}$

C
$\frac{3}{16} E_{2}$

D
$\frac{16}{3} E_{2}$

Solution

${(E_{H})}_{n} = \frac{- 13.6}{n^{2}} e V_{1}$
${(E_{H})}_{2} = E_{2} = \frac{- 13.6}{4} e V$
$E_{H e^{+}} = \frac{- Z^{2} \times 13.6}{n^{2}} e V$
${(E_{H e^{+}})}_{3} = \frac{- 4 \times 13.6}{9} e V$
$\frac{E_{H e^{+}}}{E_{2}} = \frac{16}{9}$
Question 9
1 / -0

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

A
I

B
II

C
III

D
IV

Solution

$E^{\alpha}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
$E_4 \rightarrow 2 \propto\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right) \propto\left(\frac{1}{4}-\frac{1}{16}\right) \propto \frac{3}{16}$
$E_{2} \rightarrow 1 \propto\left(\frac{1}{1}-\frac{1}{4}\right) \propto \frac{3}{4}$
$E_{2} \rightarrow_{3} \propto\left(\frac{1}{9}-\frac{1}{16}\right) \propto \frac{7}{144}$
$\mathrm{So}, \mathrm{E}_{2} \rightarrow 1$ is the highest.
Question 10
1 / -0

The wavelength of radiations emitted is $λ_{0}$ when an electron in a hydrogen atom jumps from the third orbit to the second orbit. If in the $H$ -atom itself, the electron jumps from the fourth orbit to the second orbit, then the wavelength of emitted radiation will be:

A
$\frac{20}{27} λ_{0}$

B
$\frac{16}{25} λ_{0}$

C
$\frac{27}{20} λ_{0}$

D
$\frac{29}{16} λ_{0}$

Solution

As in both cases, the electron came to the second orbit, we will consider the Balmer series,
Now, for Balmer series, the wavelength is given by,
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}}), n = 3, 4, 5$
In case 1, when $n = 3, λ = λ_{0}$
$\frac{1}{λ_{0}} = R (\frac{1}{4} - \frac{1}{9}), n = 3, 4, 5$
$R = \frac{36}{5 λ_{0}}$ ...(1)
In case 2, when $n = 4, λ$ is given by
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{4^{2}})$
$\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{16}) = \frac{27}{20 λ_{0}}$
$λ = \frac{20 λ_{0}}{27}$

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Physics Test - 3

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Physics Test - 3

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Question 1

ℓH>ℓLiand|EH|>|ELi|

ℓH=ℓLiand|EH|<|ELi|

ℓH=ℓLiand|EH|>|ELi|

ℓH<ℓLiand|EH|<|ELi|

Solution

Question 2

95R

365R

185R

4R

Solution

Question 3

\(\lambda_{1}=\lambda_{2}=4 \lambda_{3}=9 \lambda_{4}\)

\(\lambda_{1}=2 \lambda_{2}=3 \lambda_{3}=4 \lambda_{4}\)

\(4 \lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}\)

\(\lambda_{1}=2 \lambda_{2}=2 \lambda_{3}=\lambda_{4}\)

Solution

Question 4

\(\lambda_{k}-\lambda_{c}\)decreases

\(\lambda_{k}-\lambda_{c}\)increases

\(\lambda_{k}\)increases

\(\lambda_{k}\)decreases

Solution

Question 5

\(1 : 3\)

\(1 : 30\)

\(7 : 50\)

\(7 : 108\)

Solution

Question 6

(m1+m2)2n2h22m12m22r2

n2h22m1+m2r2

2n2h2m1+m2r2

m1+m2n2h22m1m2r2

Solution

Question 7

λ1+λ2+λ3=0

λ3=λ1+λ2

λ3=λ1λ2λ1+λ2

λ3=λ1+λ22

Solution

Question 8

916E2

169E2

316E2

163E2

Solution

Question 9

I

II

III

IV

Solution

Question 10

2027λ0

1625λ0

2720λ0

2916λ0

Solution

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$ℓ_{H} > ℓ_{L i}$ and $| E_{H} | > | E_{L i} |$

$ℓ_{H} = ℓ_{L i}$ and $| E_{H} | < | E_{L i} |$

$ℓ_{H} = ℓ_{L i}$ and $| E_{H} | > | E_{L i} |$

$ℓ_{H} < ℓ_{L i}$ and $| E_{H} | < | E_{L i} |$

$\frac{9}{5 R}$

$\frac{36}{5 R}$

$\frac{18}{5 R}$

$\frac{4}{R}$

$\frac{(m_{1} + m_{2}) 2 n^{2} h^{2}}{2 m_{1} 2 m_{2} 2 r^{2}}$

$\frac{n^{2} h^{2}}{2 (m_{1} + m_{2}) r^{2}}$

$\frac{2 n^{2} h^{2}}{(m_{1} + m_{2}) r^{2}}$

$\frac{(m_{1} + m_{2}) n^{2} h^{2}}{2 m_{1} m_{2} r^{2}}$

$λ_{1} + λ_{2} + λ_{3} = 0$

$λ_{3} = λ_{1} + λ_{2}$

$λ_{3} = \frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}}$

$λ_{3} = \frac{λ_{1} + λ_{2}}{2}$

$\frac{9}{16} E_{2}$

$\frac{16}{9} E_{2}$

$\frac{3}{16} E_{2}$

$\frac{16}{3} E_{2}$

$\frac{20}{27} λ_{0}$

$\frac{16}{25} λ_{0}$

$\frac{27}{20} λ_{0}$

$\frac{29}{16} λ_{0}$