The kinetic energy of an electron in H-like atom
is 6.04 eV. Area of the third Bohr orbit to which
this electron belong is
Answers
Answer:
Atoms are the basic building blocks of matter which has mass and occupies space. They cannot be chemically subdivided by ordinary means. The word ‘atom’ is derived from the Greek word atom which means indivisible. Matter can be broken into small particles called atoms which are too small to be seen through the naked eye.
As discussed earlier, atoms are composed of three types of particles namely protons, neutrons and electrons. A major proportion of mass of an atom is due to protons and neutrons while electrons have very small mass of about (9.108 X 10-28 grams).Atoms are the basic unit of matter and it is vital to have a thorough knowledge of its components like electrons, protons and neutrons.
We have enlisted the information of various particles along with their charges in the table given below:
Particle
Charge
Mass (g)
Mass (amu)
Proton
+1
1.6727 x 10-24 g
1.007316
Neutron
0
1.6750 x 10-24 g
1.008701
Electron
-1
9.110 x 10-28 g
0.000549
We discuss some of the illustrations on the structure of an atom:
Problem 1:-
A doubly ionized Lithium atom is hydrogen like with atomic number 3.
(i) Find the wavelength of radiation required to excite the electron in Li++ from the first to the third Bohr Orbit. (Ionization energy of the hydrogen atom equals 13.6 eV).
(ii) How many spectral lines are observed in the emission spectrum of the above excited system?
Solution:-
We are required to find the energy required to excite doubly ionized lithium. We know that,
En = –13.6 Z2 / n2
Hence, the excitation energy = ΔE = E3 – E1 = –13.6 × (3)2 [1/32 – 1/12]
= +13.6 × (9) [1–1/9] = 13.6 × (9) (8/9) = 108.8 eV.
Wavelength λ = hc / ΔE = (6.63 × 10–34) (3×108) / (13.6 × 8) (1.6 × 10–19)
= (6.63 × (3) / (13.6) (8)(1.16)) 10–7
= 114.26 × 10–10 m
= 1143 A
(ii) From the excited state (E3), coming back to ground state, there can be 3C2 = 3 possible radiations.
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Problem 2:-
Radiation emitted in the transition n = 3 to n = 2 orbit in a hydrogen atom falls on a metal to produce photoelectrons, the electrons emitted from the metal surface with the maximum kinetic energy are allowed to pass through a magnetic field of strength 3.125x10-4 T and it is observed that the paths have radius of curvature as 10 mm. Find,
(a) The kinetic energy of the fastest photo electrons.
(b) The work function of the metal.
(c) The wavelength of the radiation.
Solution:-
First we discuss how to find the kinetic energy of the fastest photoelectron. The formula used for finding the same is
ΔE = hv23 – 13.6 (1/22 – 1/32) eV
The Kinetic energy of the fastest photoelectrons is given by
mv2 / r = Bev
or mv = p = Ber
Kinetic Energy = 1/2 mv2 = B2e2r2 / 2m
= (3.125 × 10–4)2 × (1.6 ×)
= 0.86 eV
(a) The kinetic energy of the fastest electron is 0.86 eV
(b) The work function, Φ is given by
Φ = (1.9 – 0.86) eV = 1.04 eV
(c) The wavelength of the emitted radiation is,
λ = hc/E = (6.63 × 10–34) × (3×10–8) / 1.3 × 1.6 × 1–19 m
= 6.54 × 10–7 m
= 6540Å
Orbital velocity of electron:
(a) it is smaller in outer orbits vn = v1/n.
(b) In a particular order it is greater for heavier elements.
Radius of orbit:
(a) It varies directly as square of order of orbit r∝n2.
(b) For a particular order it is smaller for heavier elements.
Energy of electron:
(a) Electrons in the outer, electrons in higher elements are more energetic.
(b) For a particular order, electrons in higher elements are less energetic.
Wave number/wave length of emitted radiation:
(a) it depends upon the orders of two orbits between which the transition takes place.
(b) For a particular transition radiation emitted by a heavier element possesses greater wave number and hence the smaller wavelength.
Laymen series lies in ultraviolet region, Balmer series in visible region while Paschen, Bracket and P fund seri
Explanation:
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Answer:
The Answer =17.8 x 10^(-16)
Explanation:
Mark as Brainlest Answer
We know,
En=-13.6 × Z²/n² ................[En=Kinetic Energy=6.04 ev]
6.04=13.6 × Z²/ 3².............[Third Bohr orbit Given]
6.04=13.6 × Z²/9
6.04 × 9= 13.6 Z²
54.37= 13.6 Z²
54.37/13.6= Z²
4=Z²
Z=2
This Means Atom is helium[Z=2]
Area=πr²
∴Radius =?
Radius =0.529×10^-10 n²/Z.........[n=3; Z=2]
r=0.529 ×9/2×10^-10
r=2.38 ×10^-10
Now,
Area= πr²
=3.14×(2.38×10^-10)²
=17.8 × 10^-16