Chapter 2 – Atomic Structure
2×10^8 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 3.0 cm.
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Nov
2×10^8 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 3.0 cm. 2×10^8 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 3.0 cm. November 12, 2020 Category: Chapter 2 - [...]
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across the length of scale of length 20 cm long.
12
Nov
If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across the length of scale of length 20 cm long. calculate the number of carbon atoms which can be placed side by side in a straight line across [...]
Calculate the energy required for the process , He+(g)→He2+(g)+e The ionization energy for the H-atom in the grounds state is 2.18×10^−18J atom^−1.
12
Nov
Calculate the energy required for the process , He+(g)→He2+(g)+e The ionization energy for the H-atom in the grounds state is 2.18×10^−18J atom^−1. Calculate the energy required for the process He+(g)→He2+(g)+e The ionization energy for the H-atom in the grounds state is 2.18×10^−18J atom^−1. November 12, 2020 Category: Chapter 2 - Atomic Structure , Chemistry , [...]
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n=4 to n=2 of He⊕ spectrum?
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Nov
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n=4 to n=2 of He⊕ spectrum? What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n=4 to n=2 of He⊕ spectrum? November 12, 2020 Category: Chapter 2 - Atomic Structure , Chemistry , NCERT [...]
How many electron in an atom may have the following quantum number ? (a) n=4,ms=−1/2 (b) n=3,l=0
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Nov
How many electron in an atom may have the following quantum number ? (a) n=4,ms=−1/2 (b) n=3,l=0 How many electron in an atom may have the following quantum number ? (a) n=4 l=0 ms=−1/2 (b) n=3 November 12, 2020 Category: Chapter 2 - Atomic Structure , Chemistry , NCERT Exemplar Class 11 ,
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
12
Nov
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving [...]
(i) An atomic orbital has n=3. What are the possible values of l and ml ? (ii) List the quantum numbers (ml and l) of electrons for 3d-orbital. (iii) Which of the following orbitals are possible ? 1p,2s,2pand3f.
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Nov
(i) An atomic orbital has n=3. What are the possible values of l and ml ? (ii) List the quantum numbers (ml and l) of electrons for 3d-orbital. (iii) Which of the following orbitals are possible ? 1p,2s,2pand3f. (i) An atomic orbital has n=3. What are the possible values of l and ml ? (ii) [...]
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
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Nov
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron. An electron is in one of the 3d orbitals. Give the possible values of n l and ml for this electron. November 12, 2020 Category: Chapter 2 - Atomic Structure , Chemistry , NCERT [...]
What is the lowest value of n that allow g orbitals to exist?
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Nov
What is the lowest value of n that allow g orbitals to exist? What is the lowest value of n that allow g orbitals to exist? November 12, 2020 Category: Chapter 2 - Atomic Structure , Chemistry , NCERT Exemplar Class 11 ,
The mass of an electron is 9.1×10^−31 kg. If its K.E. is 3.0 × 10^−25 J, calculate its wavelength.
12
Nov
The mass of an electron is 9.1×10^−31 kg. If its K.E. is 3.0 × 10^−25 J, calculate its wavelength. calculate its wavelength. The mass of an electron is 9.1×10^−31 kg. If its K.E. is 3.0 × 10^−25 J November 12, 2020 Category: Chapter 2 - Atomic Structure , Chemistry , NCERT Exemplar Class 11 ,