The electron in a given for orbit has total energy of -1.5 electron volt calculate its kinetic energy potential energy
Answers
Answer:
ENERGY OF ELECTRON IN BOHR'S ORBIT
IIT JEE - NEET
The energy of an electron in the first Bohr orbit of H atom is -13.6 eV. The possible energy value(s) of the excited state(s) for electrons in Bohr orbits of hydrogen is (are) :
(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:
En = -13.6/n2 eV
Solution:
The energies of electrons in the Bohr's orbits of hydrogen atom expressed in eV are:
Orbit Energy
1 -13.6/12 = -13.6 eV
2 -13.6/22 = -3.4 eV
3 -13.6/32 = -1.51 eV
4 -13.6/42 = -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:
E1 : E2 : E3 : E4 ........... = 1/12 : 1/22 : 1/32 : 1/42 .......... = 1 : 1/4 : 1/9 : 1/16 ..........
Conclusion:
Correct option is "a".
Related questions
1) With increase in principal quantum number, n, the energy difference between adjacent energy levels in Hydrogen atom:
(Eamcet - 2009-M)
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)
2) If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is:
(Eamcet - 2003-E)
a) -e2/r
b) -e2/r2
c) -e2/2r
d) -e2/2r2
Solution:
Total Energy of electron, Etotal = 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 = -Ze2/r
KE = Ze2/2r
Hence:
Etotal = (-Ze2/r) + (Ze2/2r) = -Ze2/2r
And for H atom, Z = 1
Therefore:
Etotal = -e2/2r
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 nth orbit and not just for 1st 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.
3) The ionization enthalpy of hydrogen atom is 1.312 x 10-6 J mol-1. The energy required to excite the electron in the atom from n=1 to n=2 is:
(AIEEE-2008)
a) 8.51 x 105 J mol-1
b) 6.56 x 105 J mol-1
c) 7.56 x 105 J mol-1
d) 9.84 x 105 J mol-1
Logic:
We know that energy of nth orbit,
En = -K/n2 (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) = E2- E1
= (-K/n22) - (-K/n12)
= K(1/n12 - 1/n22)
= K(1/12 - 1/22)
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∞- E1
ΔE(∞,1) = (-K/n∞2) - (-K/n12) = (-K/∞2) - (-K/12) = K/n12 = 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(1/12 - 1/22)
= 1.312 x 10-6 J mol-1 (1/12 - 1/22)
= 9.84 x 105 J mol-1
Note:
Ionization energy = ΔE(∞,1) = E∞- E1 = - E1
Hence we can take negative of ionization energy as the energy of the ground state (n=1).
4) Ionization energy of He+ is 19.6 x 10-18 J atom-1. The energy of the first stationary state ( n = 1) of Li2+ is:
(AIEEE-2003)
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.
Explanation:
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