Atoms Test

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Atoms Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

A singly ionized helium atom in an excited state $$\left( n=4 \right)$$ emits a photon of energy $$2.6 eV$$. Given that the ground state energy of hydrogen atom is $$-13.6 eV$$, the energy $$\left( { E }_{ f } \right) $$ and quantum number $$\left( n \right) $$ of the resulting state are respectively,

A
$${ E }_{ f }=-13.6eV$$, $$n=1$$

B
$${ E }_{ f }=-6.0eV$$, $$n=3$$

C
$${ E }_{ f }=-6.0eV$$, $$n=2$$

D
$${ E }_{ f }=-13.6 eV$$, $$n=2$$

Solution

From the Bohr's theory of single electron species, we have the Energy of the state (quantum number = n) and atomic number = Z as
$$E=-13.6\frac{Z^2}{n^2}$$

Thus, Energy of $$4^{th}$$ state in Helium ion is $$E_4^{He^+} = -13.6\times \frac{2^2}{4^2}=-3.4eV$$

Energy of the emitted photon is $$E_p = 2.6eV$$

Energy after emitting the photon is $$E_f = E_4^{He^+}-E_p = -3.4-2.6=-6eV$$

The quantum number is given by $$n = (\frac{-13.6\times Z^2}{E_n})^{\frac{1}{2}} = \sqrt{\frac{-13.6\times 4}{-6}}=\sqrt{9}=3$$
Question 2
1 / -0

The ionization energy of $${ Li }^{ 2+ }$$ is equal to

A
$$9 hcR$$

B
$$6 hcR$$

C
$$2 hcR$$

D
$$hcR$$

Solution

Ionization energy of an atom $$=hc R{ Z }^{ 2 }$$
where $$R$$ is the Rydberg constant and $$Z$$ is the atomic number.
We know, $$Z=3$$ for $${ Li }^{ 2+ }$$
Ionization energy of $$Li^{2+}$$ $$={ \left( 3 \right) }^{ 2 }hcR$$ $$=9 hcR$$
Question 3
1 / -0

An electron in hydrogen atom after absorbing an energy photon jumps from energy state $${n}_{1}$$ to $${n}_{2}$$. Then it returns to ground state after emitting six different wavelengths in emission spectrum. The energy of emitted photons is either equal to less than the absorbed photons. The $${n}_{1}$$ and $${n}_{2}$$ are

A
$${n}_{2}=4,{n}_{1}=3$$

B
$${n}_{2}=5,{n}_{1}=3$$

C
$${n}_{2}=4,{n}_{1}=1$$

D
$${n}_{2}=5,{n}_{1}=1$$

Solution

From $$n_2=4$$,
6 lines are obtained in emission spectrum.
Now, $$E_{4-2}=E_{absorbed}$$
$$F_{4-3}<E_{absorbed}$$
and $$E_{4-1}, E_{3-1}, E_{2-1}>E_{absorbed}$$
Hence, $$n_1=2$$ and $$n_2=4$$
Question 4
1 / -0

The transition from from the state $$n = 5$$ to $$n = 1$$ in a hydrogen atom results in $$UV$$ radiation. Infrared radiation will be obtained in the transition.

A
$$2\rightarrow 1$$

B
$$3\rightarrow 2$$

C
$$4\rightarrow 3$$

D
$$6\rightarrow 2$$

Solution

$$\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$$
for UV $$n_1=1$$ and $$n_2=5$$, $$\lambda=\dfrac{1.04}{R}$$
for infrared the wavelength must be greater the UV wavelength.
For $$n_1=3$$ and $$n_2=4$$ , $$\lambda=\dfrac{7.2}{R}$$. This is the largest wavelength we can get
Question 5
1 / -0

Light of wavelength $$ \lambda = 4000 A^o $$ and intensity $$ 100 W/m^2 $$ is incident on a plate the threshold frequency $$ 5.5 \times 10^{14} Hz. $$ Find the number of photons incident $$ m^2 $$ per sec.

A
$$ 10^{21} $$

B
$$ 3.0 \times 10^{19} $$

C
$$ 2.02 \times 10^{20} $$

D
$$ 2.02 \times 10^{21} $$

Solution

Let number of photons emitted per second be $$ n. $$
Then intensity $$ = 100 W/m^2 $$
$$ \dfrac {nhc}{\lambda} = 100 $$
$$ n = \dfrac {100 \times \lambda }{hc} $$
$$ = \dfrac {100 \times 4000 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^8 } $$
$$ = 2.02 \times 10^{20} $$
Question 6
1 / -0

A gas of monoatomic hydrogen is bombarded with a stream of electrons that have been accelerated from rest through a potential difference of $$12.75V$$. In the emission spectrum one cannot observe any line of

A
Lyman series

B
Balmer series

C
Paschen series

D
Pfund series

Solution

Energy of electron stream $$=12.75eV.$$
The ground state electron having energy $$-13.6eV$$ will acquire this energy and have total energy $$=13.6+12.75=-0.85eV$$.
This corresponds to the energy of electron in the 4th orbit. So, it will radiate energy when electron jumps from 4th orbit.
Hence, Pfund series cannot be observed.
Question 7
1 / -0

The difference between a photon and neutrino is :

A
The spin of photon is 1 and that of neutrino is $$ \dfrac {1}{2} $$ in units of $$\dfrac {h}{2 \pi} $$

B
The spin of photon is $$ \dfrac {1}{2} $$ and that of neutrino is 1 in units of $$\dfrac {h}{2 \pi} $$

C
Both have equal spin but neutrino is electromagnetic in nature

D
Both have unequal spin but neutrino is electromagnetic in nature

Solution

The spin of photons is 1 and that of neutrino is $$ \dfrac {1}{2} $$ in units of $$ \dfrac {h}{2 \pi} . $$
Question 8
1 / -0

In Bohr's theory of Hydrogen atom, the electron jumps from higher orbit 'n' to lower orbit 'p'. The wavelength will be minimum for the transition

A
n = 5 to p = 4

B
n = 4 to p = 3

C
n = 3 to p = 2

D
n = 2 to p = 1

Solution

Wavelength of light emitted due to transition from $$n$$ to $$p$$ orbits $$\dfrac{1}{\lambda } = R \bigg(\dfrac{1}{p^2} - \dfrac{1}{n^2} \bigg)$$
$$\implies$$ $$\lambda = \dfrac{n^2 p^2}{R(n^2 - p^2)}$$
Case (A) : $$n =5$$ $$p =4$$
We get $$\lambda_A = \dfrac{5^2 \times 4^2}{R(5^2 - 4^2)} = 44.44 R$$
Case (B) : $$n =4$$ $$p =3$$
We get $$\lambda_B = \dfrac{4^2 \times 3^2}{R(4^2 - 3^2)} = 20.57 R$$
Case (C) : $$n =3$$ $$p =2$$
We get $$\lambda_C = \dfrac{3^2 \times 2^2}{R(3^2 - 2^2)} = 7.2 R$$
Case (D) : $$n =2$$ $$p =1$$
We get $$\lambda_D = \dfrac{2^2 \times 1^2}{R(2^2 - 1^2)} = 1.33 R$$
Thus minimum wavelength is emitted when a transition takes place from $$n=2$$ to $$p=1$$ orbit.
Question 9
1 / -0

Electrons are excited form $$n=1$$ to $$n=4$$ state. During downward transitions, possible number of spectral lines observed in Balmer series is.

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

For Balmer series it should move to n=2
so possible line are $$n=4\rightarrow n=2$$
$$n=3\rightarrow n=2$$
Question 10
1 / -0

An electron enters a parallel plate capacitor with horizontal speed $$\mu$$ and is found to deflect by angle $$\theta$$ on leaving the capacitor as shown. It is found that $$\tan\theta=0.4$$ and gravity is negligible. If the initial horizontal speed is doubled, then $$\tan$$ will be.

A
$$0.1$$

B
$$0.2$$

C
$$0.8$$

D
$$1.6$$

Solution

let the horizontal length of plate be L
now $$t=L/\mu$$
now $$v_y=at=\dfrac{eE}{m}\dfrac{L}{\mu}$$
$$tan\theta=\dfrac{v_y}{\mu}=\dfrac{eEL}{m\mu^2}=0.4$$
when speed becomes double
$$tan\theta=\dfrac{v_y}{\mu}=\dfrac{eEL}{m(2\mu^2)}=0.1$$

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Atoms Test - 43

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Atoms Test - 43

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Question 1

$${ E }_{ f }=-13.6eV$$, $$n=1$$

$${ E }_{ f }=-6.0eV$$, $$n=3$$

$${ E }_{ f }=-6.0eV$$, $$n=2$$

$${ E }_{ f }=-13.6 eV$$, $$n=2$$

Solution

Question 2

$$9 hcR$$

$$6 hcR$$

$$2 hcR$$

$$hcR$$

Solution

Question 3

$${n}_{2}=4,{n}_{1}=3$$

$${n}_{2}=5,{n}_{1}=3$$

$${n}_{2}=4,{n}_{1}=1$$

$${n}_{2}=5,{n}_{1}=1$$

Solution

Question 4

$$2\rightarrow 1$$

$$3\rightarrow 2$$

$$4\rightarrow 3$$

$$6\rightarrow 2$$

Solution

Question 5

$$ 10^{21} $$

$$ 3.0 \times 10^{19} $$

$$ 2.02 \times 10^{20} $$

$$ 2.02 \times 10^{21} $$

Solution

Question 6

Lyman series

Balmer series

Paschen series

Pfund series

Solution

Question 7

The spin of photon is 1 and that of neutrino is $$ \dfrac {1}{2} $$ in units of $$\dfrac {h}{2 \pi} $$

The spin of photon is $$ \dfrac {1}{2} $$ and that of neutrino is 1 in units of $$\dfrac {h}{2 \pi} $$

Both have equal spin but neutrino is electromagnetic in nature

Both have unequal spin but neutrino is electromagnetic in nature

Solution

Question 8

n = 5 to p = 4

n = 4 to p = 3

n = 3 to p = 2

n = 2 to p = 1

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 10

$$0.1$$

$$0.2$$

$$0.8$$

$$1.6$$

Solution

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