Ncert Solutions Chemistry Class 11th
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Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 106 m s-1, calculate de Brogile wavelength associated with this electron.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 106 m s-1, calculate de Brogile wavelength associated with this electron.
As per de Brogile’s equation,
λ = h / mv = (6.626 x 10-34 kg m2 s-1) / (9.1 x 10-31 kg) x (1.6 x 106 ms-1)
= 0.455 x 10-9 m = 0.455 nm = 455 pm.
Upvote 
The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol–1, 363.7 kJ mol–1 and 296.8 kJ mol–1 respectively. What will be the order of reactivity of these halogen acids?
The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol–1, 363.7 kJ mol–1 and 296.8 kJ mol–1 respectively. What will be the order of reactivity of these halogen acids?
This is a short answer type question as classified in NCERT Exemplar
The rate-determining step involved in the reaction is
CH3-CH=CH2 + HX → CH3-CH+-CH3 + X-
So the rate-determining step depends on the bond energy of HX i.e. higher the bond energy lesser would be reactivity.
Thus the order of reac
Upvote Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Radius of orbit of H like species = (0.529 / Z) n2Å = (52.9 / Z) n2 pm
r1 = 1.3225 nm = 1322.5 pm = (52.9 / Z) n12
r2 = 211.6 pm = 211.6 pm = (52.9 / Z) n22
∴ r1 / r2 = 1322.5 / 211.6
=>n12 /n22 = 6.25
=> n1/n2= (6.25)1/2 = 2.5
=> n1 = 2.5 n2
=> 10 n1= 25 n2
=> 2 n1= 5 n2
If n1 = 2, then n2 = 5. That means
Upvote Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 1015 (Hz) [1/32 – 1 /n2]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 1015 (Hz) [1/32 – 1 /n2]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Given, ν = (3.29 x 1015 Hz) (1/32 – 1/n2)
= (3.29 x 1015 Hz) (1/32 – 1/n2)
(3 x 108 ms-1) / (1.285 x 10-6 m) = (3.29 x 1015 Hz) (1/32 – 1/n2)
2.3346 x 1014 = (3.29 x 1015 Hz) (1/32 – 1/n2)
2.3346 / 32.9 = 1/32 – 1/n2
0.071 = 1/9 – 1/n2
1/n2 = 1/9 – 0.071= 0.111 – 0.071 = 0.04
n2 = 1/ 0.04 = 25
=> n = 5
For
Upvote Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
This is a short answer type question as classified in NCERT Exemplar
The rotation around C-C single bond is possible but it is not completely free due to the torsional strain which is about 1-20 KJ mol-1.
Upvote If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 m s-1. Calculate the energy with which it is bound to the nucleus.
If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 m s-1. Calculate the energy with which it is bound to the nucleus.
λ = 150 pm, v = 1.5 x 107 m s-1
Kinetic energy, K.E. = ½ mv2 = ½ x 9.1 x 10-31 kg x (1.5 x 107 ms-1)2
= [ (9.1 x 1.5 x 1.5) / 2] x 10-31 +14
= 10.2375 x 10-17 J = 1.02375 x 10-16 J
K.E = hc/ λ = (6.626 x 10-34 kg m2 s-1) / (3 x 108 ms-1) / (1.5 x 10-10 m)
= [ (6.626 x 3) x 10-34+8+10] / 1.5 &nb
Upvote The ejection of the photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
The ejection of the photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
λ = 256.7 nm = 256.7 x 10-9 m
K.E. = 0.35 eV
E = hc/ λ = (6.626 x 10-34Js) / (3 x 108 ms-1) / (256.7 x 10-9 m)
= (6.626 x 3 x 10-17) J/ 256.7
= (6.626 x 3 x 10-17) / (256.7 x 1.602 x 10-19) eV
E = 4.83 eV
The potential applied to silver gets converted into kinetic energy of the photoelectron.
So, Kinetic e
Upvote Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?
Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?
This is a short answer type question as classified in NCERT Exemplar
Yes, it will execute geometrical isomerism as the product formed can be either cis-2-butene or trans-2-butene.
Upvote Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) Threshold wavelength and, (b) Planck’s constant.
λ (nm)
500
450
400
v × 10–5 (cm s–1)
2.55
4.35
5.35
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) Threshold wavelength and, (b) Planck’s constant.
|
λ (nm) |
500 |
450 |
400 |
|
v × 10–5 (cm s–1) |
2.55 |
4.35 |
5.35 |
Let the threshold wavelength be λ0 nm or λ0 × 10−9 m.
h (ν−ν0) = ½ mv2
hc (1/λ−1/λ0)= ½ mv2
hc [ (1/500×10−9) – (1/ λ0×10−9)] = ½ m (2.55×106)2 . (1)
Similarly,
hc [ (1/450×10−9) – (1/λ0? ×10−9? )] = ½? m (4.35×106)2 . (2)
Similarly,
hc [ (1/400×10−9) – (1/λ0? × 10−9? )] = ½? m (5.2 × 106)2 &nb
Upvote The work function for caesium atom is 1.9 eV. Calculate (a) The threshold wavelength and (b) The threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
The work function for caesium atom is 1.9 eV. Calculate (a) The threshold wavelength and (b) The threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
W0 = 1.9 eV = 1.9 x 1.602 x 10-19 J
Threshold frequency, v0 = W0 / h = (1.9 x 1.602 x 10-19 J) / (6.626 x 10-34Js)
= 0.459 x 1015 s-1 = 4.59 x 1014 s-1
Threshold wavelength? 0 = c / v0 = (3 x 108 ms-1) / (4.59 x 1014 s-1)
= 0.6536 x 10-6 m = 653.6 nm? 654 nm
Kinetic energy, E = E0 + ½ mv2
(1/2 mv2) = E
Upvote