Q: Calculate the energy in joule corresponding to light of wavelength45 nm :(Planck's constant h=6.63×10−34Js; speed of light c=3×108 ms−1 )

EE=λ hc =45×10−96.63×10−34×3×108=4.42×10−18 J

Q: What is the maximum numbers of electrons that can be associated with the following set of quantum numbers?n=3,l=1 and m=−1

n=33px or 3pyℓ=1 no of e−1=2 m=−1

Question 1

4d, 5p, 5f and 6p orbitals are arranged in the order of decreasing energy. The correct option is:

Accepted Answer

(n + l) values for, 4d = 4 + 2 = 6									5p = 5 + 1 = 6							5f = 5 + 3 = 8							6p = 6 + 1 = 7∴  Correct order of energy would be5f > 6p > 5p > 4d

Question 2

Which of the following series of transitions in the spectrum of hydrogen atom fall in visible region?

Accepted Answer

In H-spectrum, Balmer series transitions fall in visible region.

Question 3

Two electrons occupying the same orbital are distinguished by:

Accepted Answer

Electron occupying same orbital have different spin quantum number.

Question 4

What is the maximum number of orbitals that can be identified with the following quantum numbers? n=3,ℓ=1, mt​=0

Accepted Answer

n=3,ℓ=1, m=0  Orbital is 3p2​.

Question 5

Calculate the energy in joule corresponding to light of wavelength45 nm  :(Planck's constant h=6.63×10−34Js;  speed of light c=3×108 ms−1  )

Accepted Answer

E​E=λ hc ​=45×10−96.63×10−34×3×108​=4.42×10−18 J​

Question 6

What is the maximum numbers of electrons that can be associated with the following set of quantum numbers?n=3,l=1  and m=−1

Accepted Answer

n=33px​ or 3py​​ℓ=1 no of e−1=2​ m=−1

Question 7

Based on equation E=−2.178×10−18 J(n2z2​) , certain conclusions are written. Which of them is not correct?

Accepted Answer

As we move nearer to the nucleus e−are more strongly bonded.

Question 8

Which one is the wrong statement?

Accepted Answer

Energy of 2 s-orbital and 2 p-orbital in case of hydrogen like atoms is equal.

Question 9

(i) Calculate the number of electrons which will together weighs one gram. (ii) Calculate the mass and charge of one mole of electrons.

Accepted Answer

The number of electrons which will weigh 1 g is 9.11×10−3110−3​=1.098×1027 .Mass of one mole of electrons is 9.11×10−31×6.023×1023=5.486×10−7  kg.Charge of 1 mole of electrons is 1.602×10−19×6.023×1023=9.647×104 C.

Question 10

(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total  mass of neutrons in 7 mg of 14C.(Assume that mass of a neutron =1.675×10−27kg).(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3​ at STP.Will the answer change if the temperature and pressure are changed?

Accepted Answer

(i) Total number of electrons present in 1 methane molecule  =6+4=10.Total number of electrons resent in 1 mole of methane  =10×6.023×1023=6.023×1024 electrons.The number of neutrons is equal to the difference in the mass number and atomic number.It is equal to 14−6=8.(ii)(a) The number of neutrons present in 7 mg of C-14 is 146.023×1023×8​×7×1010−3=2.4088×1021 neutrons.(b) Total mass of neutrons in 7 mg of C-14 is 2.4088 ×1021×1.675×10−274.035×10−6  kg.(iii)(a)Number of protons present in 1 molecule of ammonia =7+3=10.Number of molecules present in 34 mg of ammonia is 176.023×1023×34×10−3​=1.2044×1021.Total number of protons present in 34 mg of ammonia=10×1.2044×1021=1.2044×1022(b) Total mass of protons present in 34 mg of ammonia at STP is 1.6726×10−27×1.204×1022=2.0138×10−5 kg.When temperature and pressure are changed, there is no difference in the answer.

Question 11

How many neutrons and protons are there in the following nuclei? (i) 613​C (ii) 816​O (iii) 1224​Mg (iv) 2656​Fe (v) 3888​Sr

Accepted Answer

The number of neutrons and protons are shown below. NucleusAtomic number, Z Mass number, A Number of protons  = atomic number Number of neutrons  =A−Z  613​C  6 13 6 13−6=7  816​O   8 16 8  16−8=8  1224​Mg  12 24 12  24−12=12  2656​Fe  26 56 26  56−26=30  3888​Sr  38 88 38  88−38=50

Question 12

Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A). (i) Z = 17 ,  A = 35(ii) Z = 92 ,  A = 233(iii) Z =  4 ,   A =  9

Accepted Answer

(i) The atom with atomic number 17 and mass number 35 is chlorine. Its complete symbol is 1735​Cl.(ii) The atom with atomic number 92 and mass number 233 is uranium. Its complete symbol is 92233​U.(iii) The atom with atomic number 4 and mass number 9 is beryllium. Its complete symbol is 49​Be.

Question 13

Yellow light emitted from a sodium lamp has a wavelength (λ )   of 580 nm. Calculate the frequency (v)  and wavenumber  (vˉ )  of the yellow light.

Accepted Answer

The wavelength of yellow light is 580nm =  580 ×10−9m.The frequency of yellow light,  ν=λc​=580×10−9m3.0×108m/s​=5.17×1014/s.The wave number of yellow light, νˉ=λ1​=580×10−9m1​=1.72×106/m.

Question 14

Find energy of each of the photons which:(i) correspond to light of frequency 3 x 1015 Hz(ii) have wavelength of 0-50 A.

Accepted Answer

(i) The energy of photon is   E=hν=6.626×10−34Js×3×1015 m/s               =1.988×10−18J(ii) The energy of photon is E=λhc​=0.50×10−10m6.626×10−34Js×3×108​                  =3.98×10−15s

Question 15

Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0×10−10 s.

Accepted Answer

(i) The frequency of the light wave is period1​=  2.0×10−10s1​=5×109/s.(ii) The wavelength of light wave is νc​=  5×109/s3×108m/s​=6.0×10−2m.(iii) The wave number of light wave is νˉ= λ1​=  6.0×10−21​=16.66/m.

Question 16

What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Accepted Answer

The wavelength of light is 4000pm or 4000 ×10−12m  or   4000×10−9m.The energy one photon is λhc​=  4×10−96.626×10−34×3×108​=4.97×10−17.The number of photons that will provide 1 J of energy is 4.97×10−171J​J=2.01×1018.

Question 17

A photon of wavelength 4×10−7 m strikes on a metal surface, the work function of the metal is 2.13 eV. Calculate:(i) The energy of the photon (eV)(ii) The kinetic energy of the emission(iii) The velocity of the photoelectron. [1eV=1.6020×     10−19J].

Accepted Answer

(i)The energy of the photon is E=λhc​    =4.0×10−7m6.626 ×10−34Js ×3×108m/s​    =4.97×10−19J .Convert the unit into electron volts.E=1.602×10−194.97 ×10−19​=3.10eV(ii) The kinetic energy of emission is  K.E=hν−hνo​=3.10eV−2.13eV=0.97eV.(iii) The expression for the velocity of emitted photoelectron is v=m2K.E​​    =9.11×10−312×0.97×1.602×10−19​​=5.84×105m/s .

Question 18

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJmol−1.

Accepted Answer

The ionization energy of sodium will be equal to the energy possessed by the radiation of wavelength 242 nm. It is E=λhc​=  242×10−9m6.626 ×10−34Js×3×108 ​=8.21×10−19J.The ionization energy per mole is 8.21 ×10−19 ×6.023×1023=494000J=494kJ .

Question 19

A 25 watt bulb emits monochromatic yellow light of wavelength of  0.57μm. Calculate the rate of emission of quanta per second.

Accepted Answer

The energy of one photon isE= λhc​     =  0.57×10−66.626×10−34Js ×3×108m/s ​     =3.487×10−19J .25 watt corresponds to 25 J per second.The number of photons emitted per second is 3.487×10−1925​=7.17×1019/s.

Filters

Filters

class

subject

topic

concept

difficulty

Solutions for Structure of Atom Important Questions

Found the solution, but did not understand the concept?

Ask my first question