Class 11th

This is a Multiple Choice Questions as classified in NCERT Exemplar

 Ans: Option (iii)

Electrons move in a circular path of fixed energy called orbits .

As per Rutherford's α-particle scattering experiment the nucleus is surrounded by electrons that move around the nucleus with a very high-speed in circular paths called orbits. It does not mention the energy or stability of the electrons revolving around the nucleus. 

This is a Short Answer Type Questions as classified in NCERT Exemplar

SI unit of the mole is mol. The amount of a substance that contains as many particles or entities as there are atoms in exactly

12 g (0.012 kg) of the C-12  isotope is defined as a mole. One mole is defined as follows:

1 mole =  6.023 x 10²³

This is a Short Answer Type Questions as classified in NCERT Exemplar

On solving the above equation, the result is 5.4. All non-zeroes digits are significant. The significant figure is 2.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Mass of 1 mole of C-12 = 12g

1 mole contains 6.022*10²³ atoms.

Thus, mass of 6.022*10²³ atoms=12g

 Mass of 1 atom of carbon =126.022*10²³ g
                                         =1.99*10⁻²³ g

Thus, mass of one atom of C-12 is 1.99*10⁻²³ g

This is a Short Answer Type Questions as classified in NCERT Exemplar
Ans: The energy of electrons is determined by the value of n in the hydrogen atom and by n + l in the multielectron atom. Thus for a given principal quantum number the electrons of different orbitals would have different energy.

This is a Long Answer Type Questions as classified in NCERT Exemplar

In AB, 2 g of A combines with 5 g of B.

So, 4 g of A combines with 10 g of B.

In AB₂, 2g of B combines with 10 g of B, So 4g of A combines with 20 g of B.

In A₂B₃ , 4G of B combines with 5 g of B.

In A₂B₃ 4 g of B combined with 15 g of B.

So, the ratio between different masses of B which combine with fixed mass (4g) of A is 10: 20: 5: 15, that is, 2: 4: 1: 3.

Hence, the ratio is simple. Therefore, the law of multiple proportions is applicable.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: The uncertainty principle is significantly only for the microscopic particles and not for the macroscopic particles can be concluded by considering the following example. Let us consider a particle or an object of mass 1 milligram i.e. 10^-6 kg Then its uncertainty can be calculated as,

? x ? v = 6.626 10^-34 / 4x 3.14 $*$ 10⁶

= 10^-28 m^-2 s^-1

Thus, the value obtained is negligible and insignificant for the uncertainty principle to be applied to this particle.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: Given, m (mass) = 10g, v (speed) = 90 m/s and accuracy = 4%

Uncertainty in speed = $\frac{90*4}{100}=$  3.6 ms^-1

Uncertainty in position = h/4  πmΔv = 6.626 * 10^-34/4 * 3.14 * 10 * 3.6

=1.36 * 10^-33m

This is a Short Answer Type Questions as classified in NCERT Exemplar

Ans: