First of all, watch the following video on Energy of electron in Bohr's orbit.
(IIT JEE 1998)
a) -3.4 eV
b) -4.2 eV
c) -6.8 eV
d) +6.8 eV
Logic:
The energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:
Since Z = 1 for hydrogen above equation can be further simplified to:
E_{n} = -13.6/n^{2} eV
The energies of electrons in the Bohr's orbits of hydrogen atom expressed in eV are:
Orbit | Energy |
1 | -13.6/1^{2} = -13.6 eV |
2 | -13.6/2^{2} = -3.4 eV |
3 | -13.6/3^{2} = -1.51 eV |
4 | -13.6/4^{2} = -0.85 eV |
Excited state(s) represent n = 2, 3, 4 ...... (greater than 1).
Note:
The ratio of energy of electrons in the orbits of hydrogen atom is:
E_{1} : E_{2} : E_{3} : E_{4} ........... = 1/1^{2} : 1/2^{2} : 1/3^{2} : 1/4^{2} .......... = 1 : 1/4 : 1/9 : 1/16 ..........
Conclusion:
Correct option is "a".
a) Increases
b) Decreases
c) Remains constant
d) Decreases for lower values of n and increases for higher values of n
Solution:
From the previous problem, we can clearly see the decrease in energy difference between adjacent levels with increase in the principal quantum number, n. (see the table or ratio)
a) -e^{2}/r
b) -e^{2}/r^{2}
c) -e^{2}/2r
d) -e^{2}/2r^{2}
Solution:
Total Energy of electron, E_{total} = Potential energy (PE) + Kinetic energy (KE)
For an electron revolving in a circular orbit of radius, r around a nucleus with Z positive charge,
PE = -Ze^{2}/r
KE = Ze^{2}/2r
Hence:
E_{total} = (-Ze^{2}/r) + (Ze^{2}/2r) = -Ze^{2}/2r
And for H atom, Z = 1
Therefore:
E_{total} = -e^{2}/2r
Note:
This is not a good question because 'r' value is a variable and depends on the principal quantum number, n. Actually it is the energy of electron in the n^{th} orbit and not just for 1^{st} orbit.
Then why this question is added?
To show that the questions given in entrance exams are always not perfect. So think outside the box too.
What to do when questions like this are asked?
Be flexible in your thinking. Though questions like this are not perfect, choose the correct answers wisely among the options given.
a) 8.51 x 10^{5} J mol^{-1}
b) 6.56 x 10^{5} J mol^{-1}
c) 7.56 x 10^{5} J mol^{-1}
d) 9.84 x 10^{5} J mol^{-1}
Logic:
We know that energy of n^{th} orbit,
E_{n} = -K/n^{2} (for hydrogen atom), where K is a constant.
The energy required to excite the electron from n = 1 to n = 2 is:
ΔE_{(2,1)} = E_{2}- E_{1}
= (-K/n_{2}^{2}) - (-K/n_{1}^{2})
= K(1/n_{1}^{2} - 1/n_{2}^{2})
= K(1/1^{2} - 1/2^{2})
So we have to know the value of 'K'. This can be done by using ionization enthalpy data.
Ionization enthalpy is the energy required to take the electron from n = 1 orbit to n = ∞ orbit. Hence ionization energy must be equal to the energy difference between these two orbits.
i.e.
Ionization energy = ΔE_{(∞,1)} = E_{∞}- E_{1}
ΔE_{(∞,1)} = (-K/n_{∞}^{2}) - (-K/n_{1}^{2}) = (-K/∞^{2}) - (-K/1^{2}) = K/n_{1}^{2} = K
Therefore:
K = 1.312 x 10^{-6} J mol^{-1}
Solution:
Now we can calculate the energy required to excite the electron from n = 1 to n = 2 as follows.
ΔE_{(2,1)} = K(1/1^{2} - 1/2^{2})
= 1.312 x 10^{-6} J mol^{-1} (1/1^{2} - 1/2^{2})
= 9.84 x 10^{5} J mol^{-1}
Note:
Ionization energy = ΔE_{(∞,1)} = E_{∞}- E_{1} = - E_{1}
Hence we can take negative of ionization energy as the energy of the ground state (n=1).
1) 4.41 x 10^{-16} J atom^{-1}
2) -4.41 x 10^{-17} J atom^{-1}
3) -2.2 x 10^{-15} J atom^{-1}
4) 88.2 x 10^{-17} J atom^{-1}
Logic:
Since negative of Ionization energy is the energy of first stationery state, for He^{+}, the energy of 1st level is -19.6 x 10^{-18} J atom^{-1}.
The energy of electron in 1st level for He^{+ }can be written as:
E_{1} = -K(Z^{2}/n^{2}) = -K x (2^{2}/1^{2}) = -4K
Hence
K = -19.6 x 10^{-18} / 4 = 4.9 x 10^{-18} J atom^{-1}
Solution:
The energy of first energy level of Li^{2+} = -K(Z^{2}/n^{2}) = -K x (3^{2}/1^{2}) = -9K = -9 x 4.9 x 10^{-18} = -4.41 x 10^{-17 }J atom^{-1}
1) 1.216 x 10^{-7}m
b) 2.816 x 10^{-7}m
c) 6.500 x 10^{-7}m
d) 8.500 x 10^{-7}m
Logic:
Energy required to excite electron from n=1 to n=2 will be equal to the energy difference between these levels.
i.e.
ΔE_{(2,1)} = E_{2}- E_{1}
= (-K/n_{2}^{2}) - (-K/n_{1}^{2}) (for H atom, Z = 1)
= K(1/n_{1}^{2} - 1/n_{2}^{2})
= K(1/1^{2} - 1/2^{2})
= 3K/4
= (3/4) x 2.178 x 10^{-18} J
= 1.6335 x 10^{-18} J
Solution:
The wavelength corresponding to above excitation:
λ = hc/E = (6.626 x 10^{-34} J s x 3.0 x 10^{8} m s^{-1})_{ }/_{ }(1.6335 x 10^{-18} J )
= 1.216 x 10^{-7}m
(IIT JEE ADVANCED - 2013 )
a) Larger the value of n, the larger is the radius of orbit.
b) Equation can be used to calculate the change in energy when the electron changes orbit.
c) For n=1, the electron has a more negative energy than it does n=6 which means that the electron is more loosely bound in the smallest allowed orbit.
d) The negative sign in equation simply means that the energy for electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
Answer:
option "c" is the wrong statement.
Explanation:
The energy of electron in a particular orbit is equal to the loss in energy of electron when it is taken from infinite orbit to that orbit. Energy of electron in the infinite orbit is zero which also indicates there is no attraction between nucleus and electron. But when it is bring closer towards the nucleus, there is loss of energy due to attraction and hence the energy in the orbitals for which n < ∞ is always negative. Greater the negative value greater is the attraction.
Hence the last half part of statement given in the option "c" is wrong.
1) 1
2) 25
3) 5
4) 36
Answer:
2
Solution:
First of all we have to find the n value for the energy level.
For one electron atomic system (hydrogen like atom), the orbitals in a given orbit are degenerate irrespective of their azimuthal quantum number (l). The energy is decided by principal quantum number (n) only. Hence the number of degenerate orbitals is equal to the number of orbitals in a principal quantum level and is given by n^{2}.
Therefore, the degeneracy for the 5th level in hydrogen like atom (or ion) = 5^{2} = 25.
(new) Click here to see 3d Interactive Solved Question paper
1) Write the values of energy of ground state in hydrogen atom in different units.
2) What is the ratio of energies of electrons in the ground states of H, He^{+}, Li^{2+}and Be^{3+}?
3) Calculate the atomic number of hydrogen like species which can be ionized by an electron moving with a velocity, v = 6.56 × 10^{6} m s^{-1} .
4) Write the formula/expression for energy of electron in the n^{th} orbit of hydrogen atom.
5) What is the kinetic energy of n^{th} orbit of hydrogen atom.
6) How do you calculate the total energy of electron in the n^{th} stationary orbit of hydrogen atom?
7) The energy of an electron in the nth Bohr's orbit is proportional to ____________ .
8) What is the energy of electron in 3rd Bohr's orbit of hydrogen atom?
9) Kinetic energy (KE) of electron in a particular orbit is 3.4 eV. The potential energy is ___________ .
10) If energy of electron in a hydrogen atom is -R_{H} /9. The possible number of orbitals in this hydrogen atom is _______ . Where R_{H} is Rydberg constant.
11) What is the difference in the energies of 1st and 2nd Bohr's orbits of H-atom?
12) What is the total energy electron of an electron in the n=4 Bohr orbit of the hydrogen atom ?
13) Calculate the energy required to excite an electron of Hydrogen atom from first orbit to second orbit.
14) Why is the energy of electron negative in the hydrogen atom?
15) What is the potential energy of an electron present in n shell of Be^{3+} ion?
16) What is the energy possessed by an electron for n=infinity ?
17) Write the ratio of energy of the electron in ground state of hydrogen.
18) Calculate the potential energy of an electron in the first bohr orbit of Li^{2+} ion?