Structure of Atoms

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ANS – Option (iv)

The probability density of electrons for 2s orbital decreases uniformly as distance from the nucleus increases. Electrons at the 1s orbital decreases as we move far from the nucleus, however in case of 2s the probability decreases initially then it increases with the distance and thereafter at a certain point it starts decreasing with the distance.

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Ans: (ii) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²

As per the Hund's rule the half-filled and fully filled orbital leads to the extra stability due to the symmetry thus fully filled 3d and half-filled 4s is preferred.

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 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. 

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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.

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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.

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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

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Ans:

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Ans: We know that

λ= c/υ Given,

υ = 4.620 * 10¹⁴ Hz

Thus, λ= c/υ = (3.0 * 10⁸ m/s)/ (4.620*10¹⁴ Hz) = 649.4nm

This frequency falls under visible range

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Ans: The wavelength is defined as the distance between two consecutive crests or troughs of a wave, and it is denoted by  l .

l = 4*2.16 pm = 8.64 pm

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Ans: We have

λ = h/mv

Thus, the equation signifies that in order to have the same wavelength the electron should have higher velocity as the mass of the proton is higher than that of the electron