Atoms Test

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Atoms Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The velocity of a helium nucleus travelling in a curved path in a magnetic field is $$V$$. The velocity of a proton moving in the same curved path in the same magnetic field is :

A
$$V$$

B
$$4V$$

C
$$2V$$

D
$$V/2$$

Solution

Charge on helium nucleus $$=2\alpha$$

Mass of a helium nucleus $$= 4\beta$$

Where $$\alpha$$ and $$\beta$$ are charge and mass of the helium nuclei respectively.

According to question,

$$V = \dfrac{Bqr} {m} = \dfrac{2B\alpha r} {4\beta}$$ ------------------- (I)

For proton:
Velocity $$=\dfrac{B\alpha r} {\beta} = 2V$$ [From (I)]

Velocity of proton $$= 2V$$
Question 2
1 / -0

There are only three hydrogen atoms in a discharge tube. The analysis of spectrum shows that in all the hydrogen atoms, electrons are de-exciting from the fourth orbit. What should be the maximum number of spectral lines?

A
$$6$$

B
$$1$$

C
$$4$$

D
$$5$$

Solution

Since, from the $$n^{th}$$ state, the electron may go to $$(n - 1)^{th}$$ state,$$(n - 2)^{th}$$ state ,... 2nd state or 1st state. So there are (n 1) possible transitions starting from the $$n^{th}$$ state. The atoms reaching $$(n - 1)^{th}$$ state may make (n 2) different transitions. The atoms reaching $$(n - 2)^{th}$$ state may make (n 3) different transitions. Similarly for other lower states. During each transition a photon with energy $$h\nu$$ and wavelength $$\lambda$$ is emitted out. Hence, the total number of possible transitions is equal to the number of photons emitted.
The total number of possible transitions are
$$(n-1), (n-2), (n-3), ................, 3, 2, 1 = \dfrac{n(n-1)}{2}$$

Therefore, for transition of an electron from higher energy state $$n = 4$$ to lower energy state $$n = 1$$ the number of photons emitted are
$$\dfrac{n(n-1)}{2} = \dfrac{4(4-1)}{2} = \dfrac{12}{2} = 6$$

Here, the electrons of each hydrogen atom are in same state hence in transition will emit photons of equal wavelengths for each transition hence, the maximum spectral lines emitted are $$6$$.
Question 3
1 / -0

Rydberg atoms are the hydrogen atoms in higher excited states such atoms are observed in space.The orbit number for such an atom with radius about $$0.01\ mm$$ should be :

A
$$1$$

B
$$435$$

C
$$13749$$

D
$$117$$

Solution

Given the radius of the orbit of the exited electron $$=0.01mm$$
According to formula:
$$\Rightarrow$$ radius $$=0.529A^{0}(n^{2})$$
$$\Rightarrow 0.01\times 10^{-3}=0.529\times 10^{-10}\times n^{2}$$
$$\Rightarrow \dfrac{0.01\times 10^{-3}}{0.529\times 10^{10}}=n^{2}\Rightarrow n^{2}=189035.9168$$
$$\Rightarrow n=434.7826\Rightarrow n\approx 435$$
Question 4
1 / -0

In one revolution round the hydrogen nucleus, an electron makes five crests .The electron belongs to

A
$$1^{st}$$ orbit

B
$$4^{th}$$ orbit

C
$$5^{th}$$ orbit

D
$$6^{th}$$ orbit

Solution

As the interaction of wavelength must be constructive in interference. So the wavelength must be quantised.
So, $$n_{0}\lambda =2\pi r$$
where $$n$$ is the quantum state or the number of orbit i.e., 5 in this case, thus 5 walengths is the circumference of the orbit with 5 crests.
Question 5
1 / -0

The threshold wavelength for a surface having a threshold frequency of $$0.6\times 10^{15}$$ Hz in ($$\mathring A$$) is

A
$$100$$

B
$$2000$$

C
$$5000$$

D
$$400$$

Solution

$$\begin{array}\text { given, } \\\text { threshold wavelength, } \lambda_{0} =\frac{c}{f}\\c=\text { Speed of light }=3 \times 10^{8}\mathrm{~ms} \\f =\text { frequency }=0.6\times 10^{15} \\=6 \times 10^{14}\mathrm{~Hz}\end{array}$$
$$\begin{aligned}\therefore \quad \lambda_{0}&=\frac{3 \times 10^{8}}{6 \times10^{14}} \mathrm{~m} \\&=5 \times 10^{-7} \mathrm{~m} \\&=5000 \times 10^{-10} \mathrm{~m} \\&=5000 \dot A\end{aligned}$$
Question 6
1 / -0

The ratio of momenta of an electron and a $$\alpha$$ -particle which is accelerated from rest by a potential difference of $$100 V$$ is:

A
1

B
$$\displaystyle \sqrt {\dfrac{{2{m_e}}}{{{m_\alpha }}}} $$

C
$$\displaystyle \sqrt {\dfrac{{{m_e}}}{{{m_\alpha }}}} $$

D
$$\displaystyle \sqrt {\dfrac{{{m_e}}}{{2{m_\alpha }}}} $$

Solution

$$\dfrac{Pe}{Pa}=\sqrt(\dfrac{MeQe}{MaQa})$$
where,
$$Me = mass \ of \ electron$$
$$Ma = mass \ of \ alpha \ particle$$
$$Qe = charge \ of \ electron$$
$$Qa = charge \ of \ alpha \ particle$$
we know,
$$Qa = 2Qe$$
therefore
$$\dfrac{Pe}{Pa}=\sqrt(\dfrac{MeQe}{2MaQe})$$
So we get,
$$\dfrac{Pe}{Pa}=\sqrt(\dfrac{Me}{2Ma})$$.
Question 7
1 / -0

Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model ?

A
0.65 eV

B
1.9 eV

C
11.1 eV

D
13.6 eV

Solution

The energy of $$n^{th}$$ orbit of hydrogen atom is given by

$$E_n$$ = -$$\frac{13.6}{n^2}$$ $$eV$$

Therefore, $$E_1 = -13.6 eV$$

$$E_2 =- \frac{13.6}{2^2}$$ = $$-3.4 eV$$

$$E_3 = - \frac{-13.6}{3^2}$$ = $$-1.5 eV$$

$$E_4 = - \frac{13.6}{4^2}$$ = $$-0.85 eV$$

So, $$E_3- E_2 = -1.5 -(-3.4) = 1.9 eV$$

and $$ E_4- E_3 = -0.85- (-1.5) = .0.65 eV$$

Hence correct answer is $$11.1 eV$$
Question 8
1 / -0

An electron of hydrogen atom is revolving in third Bohr's orbit $$(n=3)$$. The number of revolutions it will undergo before making a transition to the second orbit $$(n=2)$$ is
(Assume the average life time of an excited state of the hydrogen atom is in the order of $$10^{-8}s$$ and Bohr radius $$=5.3 \times 10^{-12} m$$)

A
$$2.5\times 10^{6}$$ rev

B
$$3.5\times 10^{6}$$ rev

C
$$4.5\times 10^{6}$$ rev

D
$$1.5\times 10^{6}$$ rev

Solution

Angular lifetime of exited state $$=10^{-8}$$
Radius of 3rd qrbit $$=9a0$$ [where is boheis radius]
$$speed =\dfrac{2.18\times10^{6}}{3}m/s$$
$$frequency =\dfrac{2\pi(9a0)}{speed}$$
So, Numbers of revolutions $$=\dfrac{10^{-8}\times speed}{2\pi (9a0)}$$
$$=\dfrac{10^{8}\times2.18 \times 10^{6}\times10^{11}}{2\times3.14\times9\times5.3}$$ revolutions
$$\approx 2.5 \times10^{6}revolutions.$$
Question 9
1 / -0

The time taken by a photo-electron to come out after the photon strikes is approximately

A
$$10^{-6}$$ sec

B
$$10^{-4}$$ sec

C
$$10^{-10}$$ sec

D
$$10^{-16}$$ sec

Solution

The time by a photoeletron to come out after the
photon strikes is approximately $$10^{-10}$$ seconds. This is a fact.
So, the answer is option (C).
Question 10
1 / -0

If $$n_{r}$$ and $$n_{b}$$ are the number of photons of red and blue lights respectively with same energy, then

A
$$n_{r}>n_{b}$$

B
$$n_{r}< n_{b}$$

C
$$n_{r}= n_{b}$$

D
no relation between $$n_{r}$$ and $$n_{b}$$

Solution

As the total energy is same, frequency of red is less as compared to that of blue.
Therefore, as $$E =hv$$ (energy of a photon), energy of a photon of blue has more energy as compared to energy of photon of red.
As, total energy is same, number of red photons $$>$$ number of blue phtons
$$n_{r}>n_{b}.$$

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Atoms Test - 27

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Atoms Test - 27

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Question 1

$$V$$

$$4V$$

$$2V$$

$$V/2$$

Solution

Question 2

$$6$$

$$1$$

$$4$$

$$5$$

Solution

Question 3

$$1$$

$$435$$

$$13749$$

$$117$$

Solution

Question 4

$$1^{st}$$ orbit

$$4^{th}$$ orbit

$$5^{th}$$ orbit

$$6^{th}$$ orbit

Solution

Question 5

$$100$$

$$2000$$

$$5000$$

$$400$$

Solution

Question 6

1

$$\displaystyle \sqrt {\dfrac{{2{m_e}}}{{{m_\alpha }}}} $$

$$\displaystyle \sqrt {\dfrac{{{m_e}}}{{{m_\alpha }}}} $$

$$\displaystyle \sqrt {\dfrac{{{m_e}}}{{2{m_\alpha }}}} $$

Solution

Question 7

0.65 eV

1.9 eV

11.1 eV

13.6 eV

Solution

Question 8

$$2.5\times 10^{6}$$ rev

$$3.5\times 10^{6}$$ rev

$$4.5\times 10^{6}$$ rev

$$1.5\times 10^{6}$$ rev

Solution

Question 9

$$10^{-6}$$ sec

$$10^{-4}$$ sec

$$10^{-10}$$ sec

$$10^{-16}$$ sec

Solution

Question 10

$$n_{r}>n_{b}$$

$$n_{r}< n_{b}$$

$$n_{r}= n_{b}$$

no relation between $$n_{r}$$ and $$n_{b}$$

Solution

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