Chemistry NCERT Exemplar Solutions Class 11th Chapter Two

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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

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Ans: The line spectrum associated with any element possesses lines corresponding to specific wavelengths and these lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy i.e., quantized values.

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Ans: As per de Broglie, every object in motion has a wave character i.e. every object/matter have dual nature both particle and wave nature. But the wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.

Given, m (mass) = 100g, v=100km/hr

 We have

λ = h/mv= 238.5 * 10-³⁴ m

As the wavelength associated with the cricket ball is so small that it can't be detected.

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Ans: We have V = 109677 [1/ni ² - 1/nf ² ] cm^-1

 Given, n_i = 2, n_f=4

V =109677 [1/4 - 1/16] = 20564.44 cm^-1

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Ans: 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:  

We know that energy is inversely proportional to the wavelength

Thus, the increasing order of energy is B