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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Which of the four quantum number (n, l, `m_(l), m_(s)`) determine (a) the energy of an electron in a hydrogen atom and in a many electron atom (b) the size of an orbital (c) the shape of an orbital (d) the orientation of an orbital in space ? |
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Answer» (a) Energy of electron in hydrogen atom is determined by n and in a many electron atom by n and l (b) Size of the orbital is determined by n (c) Shape of the orbital is determined by l (d) Orientation of the orbital is determined by `m_(l)` |
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| 2. |
Which of the following sate of quantum numbers is not permissible for an electron in an atom? (i) `n = 1,l = 1, m_(l) 0, m_(s) =+ 1//2` (ii) `n = 3, l = 1, m_(1) =- 2, m_(s) =- 1//2` (iii) `n =1, l = 1, m_(l) = 0, m_(s) =+ 1//2` (iv) `n = 2, l = 0, m_(l) = 0, m_(s) = 1`A. `(i),(ii),(iii),(iv)`B. `(ii),(iii),(iv)`C. `(i),(ii),(iv)`D. `(i),(iii),(iv)` |
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Answer» Correct Answer - C (i) is not permissible because the `l` quantum number is equal to `n,` it must be less than `n`. (ii) is not permissible because the magnitude of the `m_(l)` quantum number (that is, the `m_(l)` value, ignoring its sign) must not be greater than `l`. (iv) is not permissible beacuse `m_(s)`,quantum number can be only `+1//2` or `-1//2`. |
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| 3. |
Four set ups as given here were arranged to identify the gas evolved when dilute sulphuric acid was added to zinc granules. The most appropriate set up is : A. IB. IIC. IIID. IV |
| Answer» Correct Answer - D | |
| 4. |
A compound of vanadium has a magnetic moment of `1.73BM`. Work out the electronic configuration of vanadium in the compound |
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Answer» Step I. Calculation of the number of unpaired electrons. Magnetic moment `(mu)=sqrt(n(n+2)) " " ` (Here n= number of unpaired electrons) As `" " mu=1.73` BM, therefore, `1.73=sqrt(n(n+2))` `(1.73)^(2)=n(n+2) or 3=n^(2)+2n`. `n^(2)+2n-3=0 " "or (n-1)(n+3)=0`. Therefore n=1 Step II. Configuration of vanadium ion. The electronic configuration of vanadium (Z=23) is : `1s^(2) 2s^(2) 2p^(6)3s^(2)3p^(6)3d^(3)4s^(2)` Since the ion has only one unpaired electron, therefore , its configuration is : `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(1)` Thus, vanadium is present as `V^(4+)` ion or V (IV). |
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| 5. |
Name the element in each of the following cases: (i) A bivalent anion of the element having 10 electrons (ii) A trivalent cation of the element having 10 electrons. What is the relationship between the two ions called ? |
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Answer» (i) Oxygen (ii) Aluminium The two ions are called isoelectronic. `(O^(2-) = 8 + 2 = 10, Al^(3+) = 13 - 3 = 10)` |
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| 6. |
If the position of the electron is measured within an accuracy of `+- 0.002 nm`. Calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is `h//4pi_(m)xx0.05 nm`, is there any problem in defining this value. |
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Answer» `Delta x=0.002 nm = 0.002xx10^(-9)m=2.0xx10^(-12)m` `Delta x, Deltap=(h)/(4 pi) or Delta(p)=(h)/(4pi Delta x)` `Deltap=((6.626xx10^(-34)"kg" m^(2) s^(-1)))/(4xx3.142xx(2xx10^(-12)m))=2.638xx10^(-23) kg m s^(-1)` Actual momentum (p)`=(h)/(4 pi xx 0.05nm)=((6.626xx10^(-34)"kg"m^(2)s^(-1)))/(4xx3.142xx(5xx10^(-11)m))=1.055xx10^(-24) "kg" m s^(-1)` Since actual momentum is smaller than the uncertainty in measuring momentum, therefore, the momentum of electron can not be defined. |
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| 7. |
The mass of an electron is `9.1 xx 10^(-31) kg`. If its K.E. is `3.0 xx 10^(-25)J`, calculate its wavelength |
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Answer» Here, we are given: Kinetic energy, i.e., `(1)/(2) mv^(2) = 3.0 xx 10^(-25)J` `m = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) s^(-1)` `:. (1)/(2) xx (9.1 xx 10^(-31)) v^(2) = 3.0 xx 10^(25) or v^(2) = (3.0 xx 10^(-25) xx 2)/(9.1 xx 10^(-31)) = 659340 or v = 812 m s^(-1)` `:. lamda = (h)/(mv) = (6.626 xx 10^(-34) kgm^(2) s^(-1))/((9.1 xx 10^(-31) kg) xx 812 ms^(-1)) = 8967 xx 10^(-10) m = 896.7 nm` Alternatively, K.E. `= (1)/(2) mv^(2) or v = sqrt((2K.E.)/(m))` `lamda = (h)/(mv) = (h)/(m) xx sqrt((m)/(2K.E.)) = (h)/(sqrt(2m xx K.E.)) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(sqrt(2 xx 9.1 xx 10^(-31) kg xx 3.0 xx 10^(-25) j)) = 8967 xx 10^(-10)m` |
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| 8. |
The kinetic energy of an electron is `4.55 xx 10^(-25)J`. The mass of electron is ` 9.1 xx 10^(-31)kg`. Calculate velocity, momentum and wavelength of the electron. |
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Answer» Step I. Calculation of the velocity of electron Kinetic energy `=1//2 m upsilon^(2)=4.55 xx 10^(-25)J=4.55 xx 10^(-25)kg m^(2)s^(-2)` or ` " " upsilon^(2)=(2xxKE)/(m)=(2xx(4.55 xx 10^(-25) kgm^(2)s^(-2)))/((9.1xx10^(10^(-31)kg)))=10^(6)m^(2)s^(-2)` or `" " ` Velocity`(upsilon)=(10^(6)m^(2)s^(-2))^(1//2)=10^(3) m s^(-1)`. Step II. Calculation of momentum of the electron Momentum of electron `=m upsilon =(9.1 xx 10^(-31) kg)xx(10^(3)ms^(-1))=9.1xx10^(-28)kg ms^(-1)` Step III . Calculation of wavelength of the electron According to de Broglie equation : `lambda =(h)/(m upsilon)= ((6.626 xx 10^(-34)kgm^(2) s^(-1)))/((9.1xx10^(-31)kg)xx(10^(3)m s^(-1)))` `=0.728 xx 10^(-6)m=7.28 xx 10^(-7)m` |
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| 9. |
Calculate the wavelength associated with an electron (mass `9.1 xx 10^(-31)kg`) moving with a velocity of `10^(3) m sec^(-1) (h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))` |
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Answer» Here, we are given: `m = 9.1 xx 10^(-31) kg, v = 10^(3) m sec^(-1), h = 6.6 xx 10^(-34) kg m^(2) s^(-1)` `lamda = (h)/(mv) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/((9.1 xx 10^(-31) kg) xx (10^(3) ms^(-1))) = 7.25 xx 10^(-7) m` |
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| 10. |
What must be the velocity of a beam of electrons if they are to disply a de Broglie wavelength of `100 Å` (mass of electron `= 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s`)? |
| Answer» `7.25 xx 10^(4) ms^(-1)` | |
| 11. |
A moving electron has `4.55 xx 10^(-25)` joules of kinetic energy. Calculate its wavelength (mass `= 9.1 xx 10^(-31) kg and h = 6.6 xx 10^(-34) kg m^(2) s^(-1)`) |
| Answer» Correct Answer - `7.25 xx 10^(-7) m` | |
| 12. |
Calculate the wavelength of an electron moving at `3.0 xx 10^(10) cm sec^(-1)` (mass of the electron `= 9.11 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) sec^(-1)`) |
| Answer» `2.41 xx 10^(-12) m` | |
| 13. |
Heisenberg uncertainty principle has no significance in our every day life. Explain. |
| Answer» In our daily life we can see only the moving macro and semi-micro objects. The mass of the striking photons of light is too small to cause any shift in their position by the time the reflected photons form the image of the object. For such particles, there is no problem in measuring the exact position and exact momentum simultaneously. Thus, the principle has no relevance for such objects. | |
| 14. |
i. Write the electronic conifigurations of the following ions: a. `H^(Θ)`, b. `Na^(o+)`, c. `O^(2-)`, d. `F^(Θ)` ii. What are the atomic numbers of elements whose outermost electrons are represented by a. `3s^(1)`, b. `2p^(3)`, c. `3p^(5)`? iii. Which atoms are indicated by the following configurations? a. `[He]2s^(1)`, b. `[Ne]3s^(2) 3p^(3)`, c. `[Ar]4s^(2) 3d^(1)` |
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Answer» (i) (a) `1s^(2), (b)1s^(2)2s^(2) 2p^(6) , (c) 1s^(2) 2s^(2)2p^(6), (d) 1s^(2)2s^(2)2p^(6)`. (ii) (a) Na (Z=11) has outermost electronic configuration `=3s^(1)` (b) N(Z=7)has outermost electronic configuration `=2p^(3)` (c) Fe(Z=26)has outermost electronic configuration `=3d^(6)` (iii) (a) Li (b) P (c) Sc |
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| 15. |
How fast is an electron moving if it has a wavelength equal to the distance it travels in one second? (mass of electron `= 9.1 xx 10^(-31)kg`) |
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Answer» Here `v = lamda`. Hence, `lamda = (h)/(mv) " gives" v^(2) = (h)/(m)` or `v = sqrt((h)/(m)) = sqrt((6.626 xx 10^(-34))/(9.1 xx 10^(-31))) = 2.698 xx 10^(-2) ms^(-1)` |
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| 16. |
The mass of electron is `9.11xx10^(-31)`kg. Calculate the uncertainty in its velocity if the uncertainty in position is the uncertainty in position is of the order of `+-10`pm. `(h=6.6xx10^(-34)"kg"m^(2)s^(-1))`. |
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Answer» Correct Answer - `5.76xx10^(6)m s^(-1)` According to uncertainty principle, `Deltax.*m delta v=(h)/(4pi) or Deltav=(h)/(4pi m Deltax)` `h=6.626xx10^(-34) "kg" m^(2)s^(-1), Delta x=10xx10^(-12)m=10^(-11)m, m=9.11xx10^(-31) `kg `:. " " Deltav=((6.626xx10^(-34)"kg"m^(2)s^(-1)))/(4xx3.143xx(9.11xx10^(-31)"kg")(10^(-11)m))=5.76xx10^(6)ms^(-1)`. |
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| 17. |
Calculate the product of uncertainty in position and velocity for an electron of mass `9.1 xx 10^(-31)kg` according to Heisenberg uncertainty principle |
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Answer» Correct Answer - `5.77 xx 10^(-5) m^(2) sec^(-1)` `Delta x. Delta p = (h)/(4pi) or Delta x. (m Delta v) = (h)/(4pi) or Delta x. Delta x = (h)/(4pi m)` |
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| 18. |
The mass of an electron is `9.11 xx 10^(-31) kg`. Calculate the uncertainty in its velocity if the uncertainty in its position is of the order of `+- 10`pm `(h = 6.6 xx 10^(-34) kg m^(2) s^(-1))` |
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Answer» Correct Answer - `5.76 xx 10^(6) m s^(-1)` `Delta x = 10` pm = `10 xx 10^(-12) m = 10^(-11) m` |
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| 19. |
If an electron is moving with a velocity `600 ms^(-1)` which is accurate upto 0.005%, then calculate the uncertainty in its position `(h = 6.63 xx 10^(-34)Js, " mass of electron" = 9.1 xx 10^(-31) kg)` |
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Answer» Velocity of the electron `= 600 ms^(-1)` Uncertainty in velocity `= (0.005)/(100) xx 600 ms^(-1) = 0.03 ms^(-1) = 3 xx 10^(-2) ms^(-1)` Now `(Delta x) (m Delta v) = (h)/(4pi)` `:. Delta x = (h)/(4pi m Delta v) = (6.63 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx 9.1 xx 10^(-31) kg xx 3 xx 10^(-2) ms^(-1)) = 1.93 xx 10^(-3)m` |
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| 20. |
Calculate the uncertainty in the position of an electron if the uncertainty in its velocity is `5.7 xx 10^(5) m//sec` (`h = 6.6 xx 10^(-34) kg m^(2) s^(-1)`, mass of the electron `= 9.1 xx 10^(-31) kg`) |
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Answer» Here, we are given: `Delta v = 5.7 xx 10^(5) m s^(-1), m = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) s^(-1)` Substiuting these values in the euqation for uncertainty principle, i.e., `Delta x xx (m xx Delta v) = (h)/(4pi)`, we have, `Delta x = (h)/(4pi xx m xx Delta v) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(4 xx (22)/(7) xx 9.1 xx 10^(-31) kg xx 5.7 xx 10^(5) ms^(-1)) = 1.0 xx 10^(-10) m` i.e., Uncertainty in position `= +- 10^(-10)m` |
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| 21. |
The uncertainty in position and velocity of a particle are: `10^(-10)m and 5.27 xx 10^(-24) m sec^(-1)` respectively. Calculate the mass of the particle `(h = 6.625 xx 10^(-34) J sec)` |
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Answer» Correct Answer - 0.1 kg `Delta x. m Delta v = (h)/(4pi) or m = (h)/(4pi. Delta x. Delta v)` |
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| 22. |
The approximate mass of an electron is `10^(-27)g`. Calculate the uncertainty in its velocity if the uncertainty in its position were of the order of `10^(-11)m (h = 6.6 xx 10^(-34) kg m^(2) sec^(-1))` |
| Answer» `5.25 xx 10^(6) ms^(-1)` | |
| 23. |
The uncertainty in the position and velocity of a particle are `10^(-10)` m and `5.27 xx 10^(-24) ms^(-1)` respectively. Calculate the mass of the particle. |
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Answer» According to uncertainty principle, `Deltax. mDelta upsilon =(h)/(4 pi ) or m=(h)/(4 pi Delta x Delta upsilon) , h=6.626 xx 10^(-34)kgm^(2)s^(-1)` `Deltax=10^(-10)m,Delta upsilon = 5.27 xx 10^(-24) ms ^(-1)` `:. " " m=((6.626 xx 10^(-34)kgm^(2)s^(-1)))/(4xx3.143 xx (10^(-10)m ) xx (5.27 xx 10^(-24)ms^(-1)))=0.1 kg` |
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| 24. |
Calculate the uncertainty in the velocity of a particle of mass `1.1xx10^(-27)` kg if the uncertainty in the uncertainty in its position is `3xx10^(-10)cm`. |
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Answer» Correct Answer - `1.6xx10^(4) m s^(-1)` According to uncertainty principle, `Deltax*m Deltav=(h)/(4pi) or Deltav=(h)/(4pimDeltax)` `Deltax=3xx10^(-10)=3xx10^(-12)m, m=1.1xx10^(-27) "kg" , pr=3.143`, `h=6.626xx10^(-34) "kg"m^(2)s^(-1)"` `Deltav=((6.626xx10^(-34)"kg"m^(2)s^(-1)))/(4xx3.143xx(1.1xx10^(-27)"kg")xx(3xx10^(-12)m))=1.6xx10^(4)ms^(-1)`. |
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| 25. |
Calculate the uncertainty in the momentum of an electron if it is confined to a linear region of length ` 1xx10^(-10)` metre. |
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Answer» According to uncertainty principle, `Deltax. Delta p=(h)/(4 pi) or Delta p=(h)/(4 pi Deltax) , Deltap=((6.626 xx10^(-34)kg m^(2)s^(-1)))/(4xx3 .143 xx (10^(-10)m))=5.27 xx 10^(-25) kg ms^(-1)` |
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| 26. |
The measurement of the electron position is associated with an uncertainty in momentum, which is equal to `1 xx 10^-18 g cm s^-1`. The uncertainty in electron velocity is (mass of an electron is `9 xx 10^-28 g`)A. `10^(9)cm s^(-1)`B. `10^(6)cm s^(-1)`C. `10^(5) cm s^(-1)`D. `10^(11) cm s^(-1)` |
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Answer» Correct Answer - A Since momenutm `(p)` is the product of mass `(m)` and velocity `(v)`, we have `Deltap = mDeltav` `(1xx10^(-18)g cm s^(-1)) = (9xx10^(-28)g) (Deltav)` or `Deltav = (1xx10^(-18)g cms^(-1))/(9xx10^(-28)g)` `= 0.11 xx 10^(10)cm s^(-1)` `= 1.1 xx10^(9) cm s^(-1)` |
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| 27. |
Two electrons present in M shell will differ inA. principal quantum numberB. azimuthal quantum numberC. magnetic quantum numberD. spin quantum number |
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Answer» Correct Answer - D For electrons present in M shell the value of other quantum numbers are same. But, the value of spin quantum number will be different. |
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| 28. |
What is the maximum number of electrons that can be placed in each shell?A. `2n^(2)`B. `n^(2)`C. `(2n)^(2)`D. `2n` |
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Answer» Correct Answer - A Each shells of principal quantum number `(n)` contains `n^(2)` orbitals. For example, if `n = 2`, then there are four `(2^(2))` orbitals: `2s, 2p_(x), 2p_(y)`, and `2p_(z)`. No more than two electrons can be placed in each orbital. Therefore, the maximum number of electrons is simply twice the number of orbitals that are amployed, i.e., `2n^(2)`. |
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| 29. |
Spin quantum number `(m_(s))` refers to the_______pssible orientations of the spin axis of an electron.A. infiniteB. sixC. fourD. two |
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Answer» Correct Answer - D Possible values of `m_(s)` are `+1//2` and `-1//2`. |
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| 30. |
Why the ball hit with a hockey by a player does not make a wave ? |
| Answer» Mass of the ball is large and so wavelength is negligible (because `lamda prop (1)/(m)`) | |
| 31. |
The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number, `m_(s) = -(1)/(2)`, is |
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Answer» Correct Answer - 9 Number of orbitals in the shell with n = 3 will be `= n^(2) = 3^(2) = 9`. Each orbital can have maximum 2 electrons one with spin `+(1)/(2)` and the other with `- (1)/(2)`. Hence, number of electrons with `m_(s) = - (1)/(2)` will be 9 |
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| 32. |
By what name are the following principles known ? (i) Electrons with the same spin quantum number cannot be present in the same atomic orbital. (ii) The wavelength associated with a moving particle is given by `lamda = h//p` (iii) Of a pair of conjugate properties, both cannot be measured precisely at the same time |
| Answer» (i) Pauli (ii) de Broglie (iii) Heisenberg. | |
| 33. |
Two values of spin quantum number i.e., +1/2 and -1/2 representA. up and down spin of the electrons respectivelyB. two quantum mechanical spin states which refer to the orientation of spin of the electronC. clockwise and anti-clockwise spin of the electrons respectivelyD. anti-clockwise and clockwise spin of the electrons respectively. |
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Answer» Correct Answer - B Spin angular momentum of the electron, a vector quantity can have two orientations relative to the chosen axis. These two orientations take the values of `+1//2` or `-1//2` and are called two spin states of the electrons. |
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| 34. |
The possible values of magnetic quantum number for p-orbital are |
| Answer» Correct Answer - B | |
| 35. |
How many orbitals are present in the subshells with (a) `n = 3, l = 2` (b) n = 4, l = 2 (c) n = 5, l = 2? |
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Answer» Correct Answer - Five orbitals in each case In each case, l = 1. Hence, `m = -2, -1, 0, + 1, + 2` i.e., 5 values which means 5 orbitals |
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| 36. |
An electron has a speed of `500 ms^(-1)` with an uncertainty of 0.02%. What is the uncertainty in locating its position ? [Give mass of electron `= 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s)` |
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Answer» Correct Answer - `5.77 xx 10^(-4) m` Uncertainty in speed `(Delta v) = (0.02)/(100) xx 500 m s^(-1) = 0.1 ms^(-1)` |
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| 37. |
A proton is accelerated to a velocity of ` 3xx 10^(7) m s^(-1)`. If the velocity can be measured with a precision of `+- 0.5%`, calculate the uncertainty in position of proton [`h = 6.6 xx 10^(-34)Js`, mass of proton `= 1.66 xx 10^(-27)kg`] |
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Answer» Correct Answer - `2.11 xx 10^(-13) m` `Delta v = (0.5)/(100) xx (3 xx 10^(7)) ms^(-1) = 1.5 xx 10^(5) m s^(-1)` `Delta x = (h)/(4pi m Delta v) = (6.6 xx 10^(-34) kg m^(2) s^(-1))/(4 xx 3.14 xx 1.66 xx 10^(-27) kg xx 1.5 xx 10^(5) ms^(-1)) = 2.11 xx 10^(-13) m` |
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| 38. |
Assertion. The 19th electron in potassium atom enters into 4 s-orbital and not the 3 d-orbital Reason. `(n +l)` rule is followed for determining the orbital of the lowest energy state.A. If both assertion and reason are true, and reason is the true explanation of the arrertionB. If both assertion and reason are true, but reason is not the true explanation of the assertionalC. If assertion is true, but reason is falseD. If both assertion and reason are false |
| Answer» Correct Answer - A | |
| 39. |
STATEMENT-1 `:` The `19^(th)` electron in potassium atom enters into 4s-orbital than in 3d-orbital. and STATEMENT-2 `:` `(n+1l)` rule is followed for determining the orbital of lowest energy state.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 isTrue, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False , Statement-2 isTrue |
| Answer» Correct Answer - 1 | |
| 40. |
Statement 1. In potassium atom, 19th electron enters 4s orbital and not 3d orbital. Statement 2. Orbital energies are compared by (n+l) rule.A. Statement-1 is true , Statement-2 is also true. Statement-2 is the correct explanation of Statement-1B. Statement-1 is true , Statement-2 is also true. Statement-2 is not correct explanation of Statement-1C. Statement -1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - A Statement 2 is the correct explanation for Statement 1. |
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| 41. |
Decrease in atomic numebr is not observed duringA. alpha emissionB. beta emissionC. positron emissionD. electron capture |
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Answer» Correct Answer - B `._(Z)^(A)X rarr ._(Z-2)^(A-4)Y+._(2)^(4)` He (alpha emission) During beta emission, neutron in the nucleus splits into protons and electron, the electron comes out of the nucleus as beta particles, leaving behind the proton leading to an increase to atomic number by one unit: `._(Z)^(A)Xrarr._(Z+1)^(A)T+._(_1)^(0)e(beta` emission) Positron (positive electorn) is an elementary particel with the same mass as the electron and an electric charge of equal magnitude but opposite sign: `._(Z)^(A)Xrarr_(Z-1)^(A)M+._(+1)^(0)e` (positron emission) During electorn capture, an electron from the inner shell enters into the nucelus and converts proton into neutron: `p+e^(-) rarr `n Thus, `._(Z)^(A)X +_(-0)^(0)e rarr _(Z-1)^(A)N` |
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| 42. |
As the nuclear charge increases form neon to calcium, the orbital energiesA. increaseB. increase rapidlyC. increase very slowlyD. fall |
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Answer» Correct Answer - D As nuclear charge increases, the electrons are attracted more strongly towards the nucleus and hence, their energies decrease (become more negative). |
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| 43. |
Akshay and Sauranh are the students of class IX. They have recently studied the chapter on structure of atom in the class. Both Akshy and Saurabh were performing some activities in the science laboratory. Akshy took a plastic comb and rubbed it in his dry hair. When he brought this plastic comb (rubbed in dry hair) near tiny pieces of paper, the comb attracted the pieces of paper towards it. Mean while, Saurabh took a glass rod and rubbed it with a piece of silk cloth. When he brought this glass rod (rubbed with silk cloth) near the tiny pieces of paper, the glass rod also attracted the pieces of paper towards it (just like the plastic comb). Akshy and Saurabh had studied an instrument called electroscope in class VIII. So, they decided to make use of a positively charged electroscope having diverged leaves (or opened up leaves) in their activity. Akshy took the plastic comb (rubbed in dry hair) and touched the metal top of positively charged electroscope with it. This made the diverged leaves of electroscope to fold up. Saurabh then tool the glass rod (rubbed with silk cloth) and touched the metal top of another positively charged electroscope with it. This made the diverged leaves of the electroscope to diverge (or open up) even more. Askshy did not understand the various conclusions which could be drawn from all these observations. Saurabh explained him everthing very clearly. (a) What conclusion can be drawn from the observation that a plastic comb rubbed in dry hair and a glass rod rubbed with silk cloth, both attract tiny pieces of paper ? (b) What do the above observations tell us about the atoms present on plastic comb and glass rod ? (c) (i) What conclusion do you get from the observation that when a plastic comb rubbed in dry hair is touched with the metal top of a positively charged electroscope, then its diverged leaves fold up ? (ii) What conclusion do you get form the observation that when a glass rod rubbed with silk cloth is touched with the metal top of a positvely charged electrosope, then its leaves diverage even more ? (d) What are the two types of electric charges present in atoms as shown by the above observations ? Name the subatomic particles which carry these charges. (e) (i) Which electric charges are agained by a plastic comb on rubbing in dry hair ? (ii) Which electric charge are lost by a glass rod on rubbing with silk cloth ? (f) What values are displayed by Saurabh in this episode ? |
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Answer» (a) It is a known fact that an electrically charged object can attract an uncharged object. So, both a plastic comb rubbed in dry hair and a glass rod rubbed with silk cloth are electrically charged (having electric charges on them) due to which they attract the uncharged pieces of paper. (b) These observations tell us that some charged particles are present in the atoms of plastic comb as well as glass rod (c) (i) This observation shows that the charge on plastic comb is opposite to that of positively charged electroscope. In other words, the plastic comb rubbed in dry hair carries negative electric charge. (ii) This observation shows that the charge on glass rod is of the same type as that of positively charged electroscope. In other words, the glass rod rubbed with silk cloth carries positive electric charge (d) The atoms contain negative electric charges as well as positive electric charges. The subatomic particles having negative charge are electrons whereas the subatomic particles having positve charge are protons. (A neutral atom contains an equal number of electrons and protons) (e) (i) The plastic comb gain negative charges (electrons) from dry hair on rubbing (ii) The glass rod loses negative charges (electrons) to silk cloth on rubbing (f) The various values displayed by Saurabh are (i) Awareness that atoms contain two types of electric charges : negative (electrons), and positive (protons), and (ii) Knowledge that only loosely held negative charges (electrons) can be transferred from atoms of one substance to another by friction during rubbing (the strongly held positive charges or protons cannot be transferred by friction during rubbing). |
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| 44. |
Naveen is a student of class IX in a city school. His uncle Ram Dev who lives in a village is not keeping good health. He has a tumour in his body. Ram Dev has come to city alongwith his son Ramesh for treatment. Naveen accompanied them to the most famous hospital for medical check-up and treatment. Whan Ram Dev told the person at the reception desk that he had a tumour, he was asked to go to the oncology department of the hospital. The special doctor (called oncologist) examined the tumour of Ram Dev carefully. He then removed some tissue from the tumour and sent it for biopsy, so as to find whether the tumour was malignant or not. The result of biopsy showed that the tumour was malignant. The doctor told Ram Dev that he had come to the hospital at the right time due to which his disease had been detected at an early stage and can be cured successfully. The doctor then recommended radiotherapy for Ram Dev. Naveen had come to know of the term ratiotherapy, Naveen could make out what disease isotopes in his class. So, as soon as doctor talked of radiotherapy, Naveen could make out what disease his uncle was suffering from. He also shared his knowledge of this disease with his uncle and his son. (a) What do you think is the disease Ram Dev is suffering from ? Define this disease (b) What are (i) tumour (ii) oncoloy (iii) oncologist, and (iv) biopsy ? (c) What is meant by saying that the tumour is malignant ? (d) What is ratiotherapy ? Explain it working briefly (e) Which ratioactive isotope is usually used in the treatment of this disease by radiotherapy ? How does it work ? (f) What values ae displayed by Naveen in this episode ? |
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Answer» (a) (i) Ram Dev is suffering from the disease called cancer (ii) Cancer is a disease caused by an uncontrolled division of abnormal cells in a part of the body. (b) (i) Tumour is a swelling of a part of the body caused by an abnormal growth of tissue. A tumour does not necessarily mean a cancer (ii) Oncology is the branch of medicine which studies and treats tumours and cancer (iii) An oncologist is a doctor who specialises in treating people with cancer (iv) Biopsy is the examination of tissues removed from any part of the body of a patient to know that presence, cause or exent of a disease (such as cancer). (c) By saying that the tumour is malignant, it means that the tumour is cancerous. If a tumour is not cancerous, it is said to be benign (d) Radiotheraphy is the treatment of cancer disease by using high energy radiations such as X-rays, gamma rays and electron beams, etc. In radiotheraphy, the high energy radiations destroy the cancer cells in the affected area of the body and stop them from growing and multiplying. (e) Cobalt-60 radioactive isotope is used in the treament of cancer by radiotheraphy. Cobalt-60 radioisotope emits high energy gamma rays. When the high energy gamma radiations are directed very carefully at the cancerous tumour in the human body, the cancerous cells get burnt. (f) The various values displayed by Naveen in this episode are (i) Awareness of cancer disease (ii) Knowledge of the treatment of cancer by radiotherapy (by using radioactive isotopes), and (iii) Helping nature. |
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| 45. |
The angular momentum of an electron in a given stationary state can be expressed as `m_(e)vr=n(h)/(2pi)`. Based on this expression an electron can move only in those orbits for which its angular momentum isA. equal to nB. integral multiple of `h/(2pi)`C. multiple of nD. equal to `h/(2pi)` only. |
| Answer» Correct Answer - B | |
| 46. |
Which of the following are true for an element ? (i) Atomic number = number of protons + number of electrons (ii) Mass number = number of protons + number of neutrons (iii) Atomic number = number of protons = number of neutrons (iv) Atomic number = number of protons = number of electronsA. I and iiB. I and iiiC. ii and iiiD. ii and iv |
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Answer» Correct Answer - d correct answer d The statement ii and iv are both correct |
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| 47. |
If number of electrons in an atom is 8 and number of protons is also 8, then (i) what is the atomic number of the atom? and (ii) what is the charge on the atoms ? |
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Answer» (a) Atomic number is equal to the number of protons in one atom. Since this atom contains 8 protons, so the atomic number is 8 (b) This atom contains an equal number of positively charged protons and negatively charged electrons (8 each), so it has no overall charge. That is, the charge on this atom is 0 (zero). |
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| 48. |
Write the complete symbol for (i) the nucleus with atomic number 56 and mass number 138 (ii) the nucleus with atomic number 26 and mass number 55 (iii) the nucleus with atomic number 4 and mass number 9. |
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Answer» (i) The element with atomic number 56 is Ba. Its symbol is `._(56)^(138)Ba` (ii) The element with atomic number 26 is Fe. Its symbol is `._(26)^(55) Fe` (iii) The element with atomic number 4 is Be. Its symbol is `._(4)^(9) Be`. |
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| 49. |
Calculate the atomic number of an element whose atomic nucleus has mass number 23 and neutrin number 12. What is the symbol of the element ? |
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Answer» We know that : Mass number = Atomic number + No. of neutrons So, 23 = Atomic number + 12 And, Atomic number = 23 - 12 = 11 The element having atomic number 11 is sodium and its symbol is Na. If, however, we indicate the atomic number and mass number also, then the symbol becomes `._(11)^(23)Na`, where 11 is the atomic number and 23 is the mass number. |
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| 50. |
What is the atomic number of the element with symbol Uus ?A. 117B. 116C. 115D. 114 |
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Answer» Correct Answer - A It is the correct answer. |
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