Atoms Test

Result

Atoms Test - 76

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of the radii of the electron orbits for the first excited state to the ground state for the hydrogen atom is:

A
2 : 1

B
4 : 1

C
8 : 1

D
16 : 1

Solution
Question 2
1 / -0

The wave number of first line of Balmer series of hydrogen atom is $$15200\ cm^{-1}$$. What is the wave number of first line of Balmer series of $$Li^{2+}$$ ion.

A
$$15200\ cm^{-1}$$

B
$$136800\ m^{-1}$$

C
$$76000\ cm^{-1}$$

D
$$13680\ cm^{-1}$$

Solution

As we know the energy of constant state for hydrogen species like $$Li^+2$$ or $$He^+$$ are multiple of integral type of $$Z^2$$ times the energy of hydrogen atom in stationary form$$,$$
Hence the wave number of first line of ion $$= 3^2 \times 15200 / cm$$
$$= 9 \times 15200 / cm = 136800 / cm$$
hence,
option $$(B)$$ is correct answer.
Question 3
1 / -0

In a Bohr atom the electron is replaced by a particle of mass 150 times the mass of the electron an the same charge, If $$a_0$$ is the radius of the first Bohr orbit of the orbital atom, then that of the new atom will be

A
$$150 a_0$$

B
$$\sqrt {150 a_0}$$

C
$$\sqrt {\dfrac {a_0}{150}}$$

D
$$\dfrac {a_0} {150}$$

Solution
Question 4
1 / -0

The ionization potential of H-atom is $$13.6$$V. The H-atoms in ground state are excited by monochromic radiations of photon energy $$12.09$$ eV. Then the number of spectral lines emitted by the excited atoms will be:

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution
Question 5
1 / -0

The wavelength of the first member of the Balmer series is $$656.3\ nm$$. The wavelength of the second line of the Lyman series is

A
$$81.02\ nm$$

B
$$121.6\ nm$$

C
$$364.8\ nm$$

D
$$729.6\ nm$$

Solution
Question 6
1 / -0

A $$12.5 eV$$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of spectral line will be emitted ?

A
$$1 $$ Lyman, $$2$$ Balmer

B
$$2$$ Lyman, $$1$$ Balmer

C
$$1$$ Lyman, $$1$$ Balmer, $$1$$ Paschen

D
$$3$$ Balmer

Solution
Question 7
1 / -0

The speed of sound in air is $$v$$. Both the source and observer are moving towards each other with equal speed $$u$$. The speed of wind is$$w$$ from source to observer. Then, the ratio $$(\cfrac{f}{{f}_{0}})$$ of the apparent frequency to the actual frequency is given by

A
$$\cfrac{v+u}{v-u}$$

B
$$\cfrac{v+w+u}{v+w-u}$$

C
$$\cfrac{v+w+u}{v-w-u}$$

D
$$\cfrac{v-w+u}{v-w-u}$$

Solution

$$\begin{array}{l} f=\left( { \dfrac { { v+u+w } }{ { v-u+w } } } \right) { f_{ 0 } } \\ \Rightarrow \dfrac { f }{ { { f_{ 0 } } } } =\left( { \dfrac { { v+u+w } }{ { v-u+w } } } \right) \\ Hence, \\ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$$
Question 8
1 / -0

In Rutherford's experiment the number of alpha particles scattered through an angle of $$60^0$$ by a silver foil is $$200$$ per minute. When the silver foil is replaced by a copper coil of the same thickness, the number of $$\alpha$$-particles scattered through an angle of $$60^0$$ per minute is -

A
$$200 \times \lgroup \dfrac {Z_{cu}}{Z_{Ag}} \rgroup $$

B
$$200 \times \lgroup \dfrac{Z_{cu}}{Z_{Ag}} \rgroup^2 $$

C
$$200 \times \lgroup \dfrac{Z_{Ag}}{Z_{Cu}} \rgroup $$

D
$$200 \times \lgroup \dfrac{Z_{Ag}}{Z_{Cu}} \rgroup^2 $$

Solution
Question 9
1 / -0

According to Moseley's law, the ratio of the slope of graph between $$\sqrt { f } $$ and $$Z$$ for $${ k }_{ \beta }$$ and $${ k }_{ \alpha }$$ is

A
$$\sqrt { \dfrac { 32 }{ 27 } } $$

B
$$\sqrt { \dfrac { 27 }{ 32 } } $$

C
$$\sqrt { \dfrac { 5 }{ 56 } } $$

D
$$\sqrt { \dfrac { 36 }{ 5 } } $$

Solution
Question 10
1 / -0

How can the brightness of the pattern on the screen of cathode ray tube be changed?

A
Changing the target

B
By changing the current on grid

C
By changing the negative potential on grid

D
Van't be changed

Solution