Atoms Test

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Atoms Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

White light is passed through sodium vapours contained in a thin walled glass flask and the transmitted light is examined with the help of a spectrometer. The spectrum so obtained is.

A
Absorption spectrum

B
Solar spectrum

C
Band spectrum

D
Continuous spectrum

Solution

By definition, absorption spectra is the spectra of the light transmitted through a material.
Question 2
1 / -0

If $$E_P$$ and $$E_k$$ represent potential energy and kinetic energy respectively, of an orbital electron, then according to Bohr's theory:

A
$$E_k= -E_p/2$$

B
$$E_k= -E_p$$

C
$$E_k= -2E_p$$

D
$$E_k= 2E_p$$

Solution

Potential energy of an orbital electron $$E_p=-\dfrac{Ze^2}{r}$$
Kinetic energy of an electron$$E_k=\dfrac{Ze^2}{2r}$$, where Z=atomic number
So, $$E_k=-\dfrac{E_p}{2}$$
Question 3
1 / -0

If the minimum energy of photons needed to produce photoelectric effect is 3eV, bombarding the photoelectric material by a number of photons of 2.5eV also one can get

A
photoelectrons of the same kinetic energy

B
photoelectrons of the higher kinetic energy

C
higher current

D
none

Solution

Work function of metal is given as $$\phi = 3 \ eV$$
Energy of incident photon $$E_{incident} = 2.5 \ eV$$
Emission of photo electrons from the metal surface takes place only when the energy of the incident photon is greater than the work function of the metal surface.
Since here, energy of incident photon is less than work function of metal surface. So emission of electron is not possible.
Question 4
1 / -0

When a beam of white light is passed through sodium vapours and then through a spectrometer, spectrum so obtained has two dark lines present in the yellow region. This spectrum is called

A
band spectrum

B
continuous spectrum

C
absorption spectrum of sodium

D
emission spectrum of sodium

Solution

By definition, absorption spectrum is the spectrum of the light transmitted through a material.
Question 5
1 / -0

In the hydrogen atom an electron is moving in nth orbit. The circumference s of the orbit and the de Broglie wavelength, of the moving electron are related by the equation

A
$$S=n \lambda$$

B
$$S=\dfrac{\lambda}{n}$$

C
$$S=\dfrac{n}{\lambda}$$

D
None of these

Solution

It is a postulate by Bohr , written as $$2 \pi r =n \lambda$$
circumference $$S=n\times wavelength$$
Question 6
1 / -0

If the electron in the hydrogen atoms is excited to n = 5 state, the number of frequencies present in the radiation emitted is :

A
4

B
5

C
8

D
10

Solution

The given excited state $$n = 5$$
Thus number of frequencies (or spectral lines) $$N = \dfrac{n(n-1)}{2}$$
$$\implies \ N = \dfrac{5\times (5-1)}{2} =10$$
Question 7
1 / -0

Radius of the second Bohr orbit of a singly ionised helium atom is

A
$$0.53 A^o$$

B
$$1.06 A^o$$

C
$$0.265 A^o$$

D
$$0.132 A^o$$

Solution

Radius of nth Bohr orbit $$r_n = \dfrac{0.529 n^2}{Z} \ A^o$$
For ionised Helium atom $$Z = 2$$
Second Bohr orbit $$n=2$$
$$\implies \ r_2 = \dfrac{0.529\times 2^2}{2} = 1.06 \ A^o$$
Question 8
1 / -0

The wave length of K-ray line of an anti cathode element of atomic number Z is nearly proportional to:

A
$$Z^2$$

B
$$(Z-1)^2$$

C
1/(Z-1)

D
1/$$(Z-1)^2$$

Solution

Frequency of K-ray $$\nu\propto (Z-1)^2$$
We know that $$\lambda = \dfrac{v}{\nu}$$
Thus $$\lambda \propto \dfrac{1}{(Z-1)^2}$$
Question 9
1 / -0

Linear momentum of an electron in Bohr orbit of H-atom (principal quantum number $$n$$) is proportional to :

A
$$\dfrac { 1 }{ { n }^{ 2 } } $$

B
$$\dfrac { 1 }{ n } $$

C
$$n$$

D
$${ n }^{ 2 }$$

Solution

The angular momentum in a Bohr orbit is given as
$$L=mvr=\dfrac{nh}{2\pi}$$
$$\implies mv=\dfrac{nh}{2\pi r}$$
$$\implies p=mv\propto n$$
Question 10
1 / -0

When electron in hydrogen atom jumps from second orbit to first orbit, the wavelength of radiation emitted is $$\lambda$$. When electron jumps from third orbit to first orbit, the wavelength of emitted radiation would be

A
$$\cfrac { 27 }{ 32 } \lambda $$

B
$$\cfrac { 32 }{ 27 } \lambda $$

C
$$\cfrac { 2 }{ 3 } \lambda $$

D
$$\cfrac { 3 }{ 2 } \lambda $$

Solution

Wavelength of photon emitted $$\dfrac{1}{\lambda} = R \bigg(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\bigg)$$ $$(\because Z = 1)$$
Transaction from second orbit to first orbit :
We have, $$n_1 = 2$$ $$n_2 = 1$$
$$\therefore$$ $$\dfrac{1}{\lambda} = R \bigg(\dfrac{1}{1^2} - \dfrac{1}{2^2}\bigg)$$ $$\implies R = \dfrac{4}{3\lambda}$$
Transaction from third orbit to first orbit :
We have, $$n_1 = 3$$ $$n_2 = 1$$
$$\therefore$$ $$\dfrac{1}{\lambda'} = R \bigg(\dfrac{1}{1^2} - \dfrac{1}{3^2}\bigg)$$
$$\implies \lambda' =\dfrac{9}{8R}= \dfrac{9}{8}\times \dfrac{3\lambda}{4} = \dfrac{27}{32}\lambda$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 41

Result

Atoms Test - 41

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SHARING IS CARING

Question 1

Absorption spectrum

Solar spectrum

Band spectrum

Continuous spectrum

Solution

Question 2

$$E_k= -E_p/2$$

$$E_k= -E_p$$

$$E_k= -2E_p$$

$$E_k= 2E_p$$

Solution

Question 3

photoelectrons of the same kinetic energy

photoelectrons of the higher kinetic energy

higher current

none

Solution

Question 4

band spectrum

continuous spectrum

absorption spectrum of sodium

emission spectrum of sodium

Solution

Question 5

$$S=n \lambda$$

$$S=\dfrac{\lambda}{n}$$

$$S=\dfrac{n}{\lambda}$$

None of these

Solution

Question 6

4

5

8

10

Solution

Question 7

$$0.53 A^o$$

$$1.06 A^o$$

$$0.265 A^o$$

$$0.132 A^o$$

Solution

Question 8

$$Z^2$$

$$(Z-1)^2$$

1/(Z-1)

1/$$(Z-1)^2$$

Solution

Question 9

$$\dfrac { 1 }{ { n }^{ 2 } } $$

$$\dfrac { 1 }{ n } $$

$$n$$

$${ n }^{ 2 }$$

Solution

Question 10

$$\cfrac { 27 }{ 32 } \lambda $$

$$\cfrac { 32 }{ 27 } \lambda $$

$$\cfrac { 2 }{ 3 } \lambda $$

$$\cfrac { 3 }{ 2 } \lambda $$

Solution

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