# Revisal Problems (Past 13 Years) JEE Advanced (Structure Of Atom)

## 29 Questions MCQ Test Chemistry for JEE Advanced | Revisal Problems (Past 13 Years) JEE Advanced (Structure Of Atom)

Description
Attempt Revisal Problems (Past 13 Years) JEE Advanced (Structure Of Atom) | 29 questions in 58 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Advanced for JEE Exam | Download free PDF with solutions
QUESTION: 1

### Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct. Q. Energy for 7.25 x 1015 photons of 5,37 x 1014 s-1 frequency in Einstein unit is

Solution:

For N photons E = Nhv
= 7.25 x 1015 x 6.62 x 10-34 Jsx 5.37 x 1014 s-1
For N0 photons with same frequency

QUESTION: 2

### What is the minimum amount of time required to measure the kinetic energy of an electron with a speed of (100 ± 1)ms-1

Solution:

Velocity varies within ±1 ms-1
Thus, it can be 99 ms-1 to 101 ms-1

QUESTION: 3

### When a certain metal was irradiated with light of frequency 1.6 x 1016 Hz, the photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency 1.0 x 1016 Hz. Thus, threshold frequency for the metal is

Solution:

For photoelectric effect

QUESTION: 4

Consider the following dissociation of O2 (dissociation energy 498 kJ mol -1)

is more energetic than normal oxygen atom (O) by 1.967 eV. Thus, maximum wavelength for photochemical dissociation is

Solution:

QUESTION: 5

Which of the following pairs of orbitals have electron density along the axis?

Solution:

Electron-density along axis.

QUESTION: 6

The normalised wave function of 1s orbital is  and the radial distribution function

Solution:

Most probable distance is when

QUESTION: 7

For any axis (say x-axis) Schrodinger wave equation is represented as

Its solution can be

Solution:

QUESTION: 8

If a 1.00 g body is travelling along the X-axis at 100 cm s -1 within 1 cm s -1, then uncertainty in its position is

Solution:

There is uncertainty with 1 cm s -1, hence, velocity varies between 99 cm s-1 to 101 cms-1.
Thus, Δv (uncertainty in velocity) = 101 - 99 = 2 cm s-1
= 0.02 m s-1
By Heisenberg's uncertainty principle

QUESTION: 9

Iodine molecule dissociates into atoms after absorbing light of . If one quantum of radiation is absorbed by each molecule, kinetic energy of each iodine atom is (Bond energy of l2 = 240kJ mol-1)

Solution:

KE given to iodine molecule = [Energy absorbed - energy required to dissociate l2]
Bond energy = 240 kJ mol-1 = 240 x 103 J mol-1

QUESTION: 10

Quantum numbers of the electrons are give

How many electrons in an atom may have these quantum numbers?

Solution:

3s-suborbit can have 2(2l + 1) = 2 electrons Only two electrons.
Ill : n = 2, l = 1
2 p-suborbit can have 2(2l + 1) = 6 electrons Thus, six electrons.
IV : n = 2, m; = 0
n = 2, l = 0 , m , = 0
n = 2, l = 1, m, = 1 - 1 , 0, + 1
Thus, four electrons.

QUESTION: 11

Direction (Q. Nos. 11-13) This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Q.

Statement I :  On heating a metal for a longer time, radiations become white and then blue as the temperature increases.
Statement II : Radiations emitted go from a lower frequency to higher frequency as the temperature increases.

Solution:

Colour of the emitted radiations change as temperature increases

Thus, as temperature changes frequency also changes and colour changes.
Thus,both Statements I and II are correct and Statement II is the correct explanation of Statement I

QUESTION: 12

Statement I : Black body is an ideal body that emits and absorbs radiations of all frequencies.

Statement II : The frequency of radiations emitted by a body goes from a lower frequency to higher frequency with an increase in temperature.

Solution:

Both Statements I and II are correct but Statement II is not the correct explanation of Statement

QUESTION: 13

Statement I : Electrons are negatively charged.

Statement II : The application of electric and magnetic field deflected the rays in the discharge tube towards the positive plate.

Solution:

Electrons are negatively charged thus, Statement I is correct. In cathode ray tube, these rays are deflected towards positive plate thus, Statement II is also correct and also the correct explanation of Statement I.

*Multiple options can be correct
QUESTION: 14

Direction (Q. Nos. 14-16) This section contains 3 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q. Spectral series of H-atom given depending on the types of transition. Which is/are correct matchings?

Solution:

*Multiple options can be correct
QUESTION: 15

Velocity of the electron in third orbit of Li2+ is equal to velocity of the electron in

Solution:

Velocity of the electron of the species with atomic number Z in nth orbit is

*Multiple options can be correct
QUESTION: 16

If the quantum num ber l could have the value of n also the

Solution:

If n = n then l = 0 ,1 ,2 , 3...... n
(a, b, c)
If n = n then l = 0 , 1 , 2 , 3 , . . . n

Both have unpaired electrons in c/-orbital thus, coloured : Correct

QUESTION: 17

Direction (Q. Nos. 17-22) This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d)

Passage I

The ionisation energy of hydrogen atom is 13.6 eV and the first ionisation energy of helium is 24.6 eV.

Q. What is the first ionisation energy of He+ (isoelectronic of H-atom)?

Solution:

QUESTION: 18

Passage I

The ionisation energy of hydrogen atom is 13.6 eV and the first ionisation energy of helium is 24.6 eV.

Q. What is the enthalpy change of the following process

Solution:

QUESTION: 19

Passage II

The position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nm

Q. Uncertainty in momentum is

Solution:

QUESTION: 20

Passage II

The position and momentum of electron of energy 0.5 keV are simultaneously determined. Its position is located within 0.2 nm

Q. Percentage uncertainty in momentum is

Solution:

QUESTION: 21

Passage III

Yellow lamp of 100 W emits radiations of wavelength 560 nm.

Q. Number of photons emitted by this lamp in 10 s is

Solution:

QUESTION: 22

Passage III

Yellow lamp of 100 W emits radiations of wavelength 560 nm.

Q. Time taken to produce 1 mole of the photons by this lamp is

Solution:

*Answer can only contain numeric values
QUESTION: 23

Direction (Q. No. 23-27) This section contains 5 question. When worked out will result in an integer from 0 fo 9 (both inclusive).

Q. What is the maximum number of emission lines when the excited electron of a H-atom in n = 4 drops to the ground state?

Solution:

*Answer can only contain numeric values
QUESTION: 24

Ionisation energy of H-atom in the ground state is 25 times the ionisation energy when it is in nth state. What is value of n?

Solution:

*Answer can only contain numeric values
QUESTION: 25

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as

Calculate value of n if, the transition is observed at 1285 nm.

Solution:

*Answer can only contain numeric values
QUESTION: 26

The atomic mass of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73°C is 'rri times that of the de-Broglie wavelength of Ne at 727°C.m i s ......

Solution:

*Answer can only contain numeric values
QUESTION: 27

The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?

Solution:

QUESTION: 28

Direction (Q. Nos. 28 and 29) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

Match the name of units given in Column I with their relation (in terms of physical constants) given in Column II.

Solution:

QUESTION: 29

Match the species in Column I with their related properties in Column II.

Solution:

Paramagnetic due to unpaired electron and sum of spin

Paramagnetic and coloured due to unpaired electrons in 3d Sum of spin= 1.5

 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code