Structure of Atom Test

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Structure of Atom Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of $$He^+$$ is :

A
$$\dfrac{36A}{7}$$

B
$$\dfrac{5A}{9}$$

C
$$\dfrac{36A}{5}$$

D
$$\dfrac{9A}{5}$$

Solution

For Lyman series $$\displaystyle hydrogen$$ atom, the shortest wavelength corresponds to:
$$n_1 = 1$$ and $$n_2 = {\infty}$$

$$\displaystyle \dfrac {1}{\lambda} = R Z^2[\dfrac {1}{n_1^2}-\dfrac {1}{n_2^2}]$$

$$\displaystyle \dfrac {1}{A} = R [\dfrac {1}{1^2}-\dfrac {1}{\infty ^2}]$$

$$\displaystyle \dfrac {1}{A} = R $$ .....(1)

For Paschen series of $$\displaystyle He^+$$ ion, the longest wavelength corresponds to :
$$n_1 = 3$$ and $$n_2 = 4$$

$$\displaystyle \dfrac {1}{\lambda} = RZ^2 [\dfrac {1}{n_1^2}-\dfrac {1}{n_2^2}]$$

$$\displaystyle \dfrac {1}{\lambda} = \dfrac {1}{A} \times 2^2 \times [\dfrac {1}{3^2}-\dfrac {1}{4^2}]$$

$$\displaystyle \lambda = \dfrac {36A}{7}$$
Question 2
1 / -0

If the principle quantum number n = 6, the correct sequence of filling of electron will be :

A
$$ns\rightarrow$$ $$(n - 2)f$$ $$\rightarrow$$ np $$\rightarrow (n - 1)$$

B
$$ns\rightarrow$$ $$(n - 1)f$$ $$\rightarrow$$ $$np$$ $$\rightarrow (n - 2)$$

C
$$ns\rightarrow np\rightarrow(n - 1)d\rightarrow(n - 2)f$$

D
$$ns\rightarrow (n - 2)f \rightarrow (n - 1)d \rightarrow np$$

Solution

If the principle quantum number n = 6, the correct sequence of filling of electron will be $$ns\rightarrow(n - 2)f\rightarrow(n - 1)d\rightarrow$$np.
Here, $$n=6$$. Hence, the correct sequence of filling of electron will be $$6s\rightarrow4f\rightarrow5d\rightarrow$$$$6p$$.
Question 3
1 / -0

Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying $$0.5\ V$$ when the radiation of $$250\ nm$$ is used. The work function of the metal is:

A
$$5.5\ eV$$

B
$$4\ eV$$

C
$$5\ eV$$

D
$$4.5\ eV$$

Solution

From Ejection photoelectron
$$\because hv = hv_o + K.E.$$
$$\dfrac{hv}{x} = \phi+\text{Stopping pot}$$
$$\therefore \phi = \dfrac{hc}{x} - S.P.$$

$$=\dfrac{6.6\times 10^{-34} \times 3\times 10^8}{250\times 10^{-9}\times 1.6\times 10^{-19}}-0.5\times 1.6\times 10^{-10}$$

$$=\dfrac{6.6\times 3}{250\times 1.6} \times 10^2-0.5$$

$$=4.95 - 0.5$$
$$\sim 4.5 eV$$

Question 4

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Directions For Questions

Answer the following by appropriately matching the lists based on the information given in the paragraph.
Consider the Bohr's model of a one -electron atom where the electron moves around the nucleus. In the following $$List-I$$ contains some quantities for the $$n^{th}$$ orbit of the atom and $$List-II$$ contains options showing how they depend in $$n$$.

	List-I		List-II
(i)	Radius of the $$n^{th}$$ orbit	(P)	$$\propto n^{-2}$$
(ii)	Angular momentum of the electron in the $$n^{th}$$ orbit	(Q)	$$\propto n^{-1}$$
(iii)	Kinetic energy of the electron in the $$n^{th}$$ orbit	(R)	$$\propto n^{0}$$
(IV)	Potential energy of the electron in the $$n^{th}$$	(S)	$$\propto n^{1}$$
		(T)	$$\propto n^{2}$$
		(U)	$$\propto n^{\frac{1}{2}}$$

...view full instructions

Which of the following has the correct combination considering $$List-I$$ and $$List-II$$?

$$(II),(R)$$

$$(II),(Q)$$

$$(I),(P)$$

$$(I),(T)$$

Solution

The kinetic energy of the electron in the $$n^{th}$$ orbit is given as:

$$K.E.=\dfrac{me^4}{8{{\epsilon}_0}^2n^2h^2}$$

$$\therefore \ K.E.\propto n^{-2}$$

Similarly,

radius of $$n^{th}$$ orbit $$r\propto \dfrac{n^2}{Z}$$

Angular momentum $$|L|\propto n$$

and Potential energy $$P.E. \propto \dfrac {-Z^2}{n^2}$$

Question 5

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Directions For Questions

	List-I		List-II
(i)	Radius of the $$n^{th}$$ orbit	(P)	$$\propto n^{-2}$$
(ii)	Angular momentum of the electron in the $$n^{th}$$ orbit	(Q)	$$\propto n^{-1}$$
(iii)	Kinetic energy of the electron in the $$n^{th}$$ orbit	(R)	$$\propto n^{0}$$
(IV)	Potential energy of the electron in the $$n^{th}$$	(S)	$$\propto n^{1}$$
		(T)	$$\propto n^{2}$$
		(U)	$$\propto n^{\frac{1}{2}}$$

...view full instructions

Which of the following has the correct combination considering List-I and List-II

$$(III),(S)$$

$$(IV),(Q)$$

$$(III),(P)$$

$$(IV),(U)$$

Solution

The kinetic energy of the electron in the $$n^{th}$$ orbit is given as:

$$K.E.=\dfrac{me^4}{8{{\epsilon}_0}^2n^2h^2}$$

$$\therefore \ K.E.\propto n^{-2}$$

Similarly,

radius of $$n^{th}$$ orbit $$\quad \propto \dfrac{n^2}{Z}$$

Angular momentum $$|L|\propto n$$

and Potential energy $$P.E. \propto \dfrac {-Z^2}{n^2}$$

Hence, the correct option is $$A$$

Question 6
1 / -0

For which one of the following, Bohr model is not valid?

A
Singly ionised helium atom $$(He^+)$$

B
Deuteron atom

C
Singly ionised neon atom $$(Ne^+)$$

D
Hydrogen atom

Solution

Bohr's model is valid only for the single electronic systems. Therefore, it will not be valid for a singly ionised neon atom as it has more electrons.
Question 7
1 / -0

The number of protons, neutrons and electrons in $$^{175}_{71}Lu$$, respectively, are

A
$$104, 71$$ and $$71$$

B
$$71, 71$$ and $$104$$

C
$$ 175, 104$$ and $$71$$

D
$$71, 104$$ and $$71$$

Solution

$$^{175}_{71}Lu$$

$$n_p= n_e=71$$

$$n_p +n_n =175$$

$$n_n=175-71=104$$

Option D is correct.
Question 8
1 / -0

Assertion: Atoms are not electrically neutral.
Reason: Number of protons and electrons are different.

A
Assertion and reason both are correct statements and reason is correct explanation for assertion.

B
Assertion and reason both are correct statements but reason is not correct explanation for assertion.

C
Assertion is correct statement but reason is wrong statement.

D
Assertion is wrong statement but reason is correct statement.

E
Both assertion and reason statements are wrong.

Solution

Assertion: Atoms are electrically neutral. They do not have net electrical charge.
Reason: Number of protons and electrons are equal. Hence, total positive charges are balanced with total negative charges.
Question 9
1 / -0

Which of the following electronic configurations has maximum energy ?

A

B

C

D

Solution

The order of increasing energy of the sub-atomic orbitals is s < p < d < f.

The energy in an excited state is more than that in the ground state.

In Option (a), two electrons are in the excited state, therefore, it has maximum energy.
Question 10
1 / -0

What would be the radius of second orbit of $${He}^{+}$$ ion?

A
$$1.058\mathring { A } $$

B
$$3.023\mathring { A } $$

C
$$2.068\mathring { A } $$

D
$$4.458\mathring { A } $$

Solution

Radius of the nth orbit, $$r_n = 0.529 \dfrac{n^2}{Z}$$ $$A^o$$
For $$He^+$$ ion, $$Z = 2$$
Radius of 2nd orbit, $$r_2 = 0.529\dfrac{2^2}{2}$$
$$\implies r_2 = 1.058$$ $$\mathring A$$

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Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 16

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Structure of Atom Test - 16

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Question 1

$$\dfrac{36A}{7}$$

$$\dfrac{5A}{9}$$

$$\dfrac{36A}{5}$$

$$\dfrac{9A}{5}$$

Solution

Question 2

$$ns\rightarrow$$ $$(n - 2)f$$ $$\rightarrow$$ np $$\rightarrow (n - 1)$$

$$ns\rightarrow$$ $$(n - 1)f$$ $$\rightarrow$$ $$np$$ $$\rightarrow (n - 2)$$

$$ns\rightarrow np\rightarrow(n - 1)d\rightarrow(n - 2)f$$

$$ns\rightarrow (n - 2)f \rightarrow (n - 1)d \rightarrow np$$

Solution

Question 3

$$5.5\ eV$$

$$4\ eV$$

$$5\ eV$$

$$4.5\ eV$$

Solution

Question 4

$$(II),(R)$$

$$(II),(Q)$$

$$(I),(P)$$

$$(I),(T)$$

Solution

Question 5

$$(III),(S)$$

$$(IV),(Q)$$

$$(III),(P)$$

$$(IV),(U)$$

Solution

Question 6

Singly ionised helium atom $$(He^+)$$

Deuteron atom

Singly ionised neon atom $$(Ne^+)$$

Hydrogen atom

Solution

Question 7

$$104, 71$$ and $$71$$

$$71, 71$$ and $$104$$

$$ 175, 104$$ and $$71$$

$$71, 104$$ and $$71$$

Solution

Question 8

Assertion and reason both are correct statements and reason is correct explanation for assertion.

Assertion and reason both are correct statements but reason is not correct explanation for assertion.

Assertion is correct statement but reason is wrong statement.

Assertion is wrong statement but reason is correct statement.

Both assertion and reason statements are wrong.

Solution

Question 9

Solution

Question 10

$$1.058\mathring { A } $$

$$3.023\mathring { A } $$

$$2.068\mathring { A } $$

$$4.458\mathring { A } $$

Solution

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