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241.
A stream of electrons from a heated filaments was passed two charged plates kept at a potential difference $$V$$ $$esu.$$ If e and m are charge and mass of an electron, respectively, then the value of $$\frac{h}{\lambda }$$ ( where $$\lambda $$ is wavelength associated with electron wave ) is given by :
A
$$\sqrt {meV} $$
B
$$\sqrt {2meV} $$
C
$$meV$$
D
$$2meV$$
Answer :
$$\sqrt {2meV} $$
As electron of charge $$'e’$$ is passed through $$'V'$$ volt, kinetic energy of electron will be $$eV$$
Wavelength of electron wave $$\left( \lambda \right) = \frac{h}{{\sqrt {2m.K.E} }}$$
$$\lambda = \frac{h}{{\sqrt {2meV} }} \Rightarrow \,\frac{h}{\lambda } = \sqrt {2meV} $$
242.
Two electrons present in $$M$$ shell will differ in
A
principal quantum number
B
azimuthal quantum number
C
magnetic quantum number
D
spin quantum number.
Answer :
spin quantum number.
For electrons present in $$M$$ shell the value of other quantum numbers are same. But, the value of spin quantum number will be different.
243.
Few electrons have following quantum numbers,
$$\eqalign{
& \left( {\text{i}} \right)n = 4,l = 1 \cr
& \left( {{\text{ii}}} \right)n = 4,l = 0 \cr
& \left( {{\text{iii}}} \right)n = 3,l = 2 \cr
& \left( {{\text{iv}}} \right)n = 3,l = 1 \cr} $$
Arrange them in the order of increasing energy from lowest to highest.
A
(iv) < (ii) < (iii) < (i)
B
(ii) < (iv) < (i) < (iii)
C
(i) < (iii) < (ii) < (iv)
D
(iii) < (i) < (iv) < (ii)
Answer :
(iv) < (ii) < (iii) < (i)
(i) 4$$p,$$ (ii) 4$$s,$$ (iii) 3$$d,$$ (iv) 3$$p,$$
The order of increasing energy
(iv) < (ii) < (iii) < (i)
244.
What is the trend of energy of Bohr's orbits ?
A
Energy of the orbit increases as we move away from the nucleus.
B
Energy of the orbit decreases as we move away from the nucleus.
C
Energy remains same as we move away from the nucleus.
D
Energy of Bohr's orbit cannot be calculated.
Answer :
Energy of the orbit increases as we move away from the nucleus.
Energy of the orbit increases as we move away from the nucleus or as the value of $$n$$ increases.
245.
For which one of the following sets of four quantum numbers, an electron will have the highest energy?
$$n$$
$$l$$
$$m$$
$$s$$
(a)
3
2
1
$$\frac{1}{2}$$
(b)
4
2
-1
$$\frac{1}{2}$$
(c)
4
1
0
$$ - \frac{1}{2}$$
(d)
5
0
0
$$ - \frac{1}{2}$$
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(b)
$$\eqalign{
& {\text{For}}\,\,n = 3,\,l = 2\,{\text{the subshell is}}\,\,3d\left( {n + l = 5} \right) \cr
& n = 4,l = 2\,{\text{the subshell is}}\,\,4d\left( {n + l = 6} \right) \cr
& n = 4,\,l = 1\,{\text{the subshell is}}\,\,4p\left( {n + l = 5} \right) \cr
& n = 5,\,l = 0,\,{\text{the subshell is}}\,\,5s\left( {n + l = 5} \right) \cr} $$
According to $$(n + l)$$ rule greater the $$(n + l)$$ value, greater the energy that is 6.
246.
Ionisation energy of $$H{e^ + }$$ is $$19.6 \times {10^{ - 18}}J\,ato{m^{ - 1}}.$$ The energy of the first stationary state $$( n = 1 )$$ of $$L{i^{2 + }}$$ is
A
$$4.41 \times {10^{ - 16}}J\,ato{m^{ - 1}}$$
B
$$ - 4.41 \times {10^{ - 17}}J\,ato{m^{ - 1}}$$
C
$$ - 2.2 \times {10^{ - 15\,}}J\,ato{m^{ - 1}}$$
247.
The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition from the
A
fourth Bohr orbit to the first Bohr orbit
B
fifth Bohr orbit to the second Bohr orbit
C
sixth Bohr orbit to the third Bohr orbit
D
seventh Bohr orbit to the third Bohr orbit.
Answer :
fifth Bohr orbit to the second Bohr orbit
In Balmer series, $${n_1} = 2$$ and $${n_2} = 3,4,5$$ So, third line arises by the emission of 5th orbit to 2nd orbit.
248.
An electron travels with a velocity of $$x\,m{s^{ - 1}}.$$ For a proton to have the same de-Broglie wavelength, the velocity will be approximately :
249.
For emission line of atomic hydrogen from $${n_i} = 8$$ to $${n_f} = n,$$ the plot of wave number $$\left( {\bar v} \right)$$ against $$\left( {\frac{1}{{{n^2}}}} \right)$$ will be (The Rydberg constant, $${R_H}$$ is in wave number unit)
A
Linear with intercept $$ - {R_H}$$
B
Non linear
C
Linear with slope $${R_H}$$
D
Linear with slope $$ - {R_H}$$
Answer :
Linear with slope $$ - {R_H}$$
$$\eqalign{
& {\text{As we know, }} \cr
& \bar v = - {R_H}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,{Z^2}\,\left( {{\text{where}},\,Z = 1} \right) \cr
& {\text{After putting the values, we get }} \cr
& \bar v = - {R_H}\left( {\frac{1}{{{n^2}}} - \frac{1}{{{8^2}}}} \right) \cr
& \therefore \,\,\bar v = \frac{{{R_H}}}{{64}} - \frac{{{R_H}}}{{{n^2}}} \cr
& {\text{Comparing to y = }}mx{\text{ + }}c{\text{, we get}} \cr
& x{\text{ = }}\,{\text{and}}\,m{\text{ = }}\, - {{\text{R}}_H}\left( {{\text{slope}}} \right) \cr} $$
250.
What is the angular velocity $$\left( \omega \right)$$ of an electron occupying second orbit of $$L{i^{2 + }}\,ion?$$
A
$$\frac{{8{\pi ^3}m{e^4}}}{{{h^3}}}{K^2}$$
B
$$\frac{{8{\pi ^3}m{e^4}}}{{9{h^3}}}{K^2}$$
C
$$\frac{{64}}{9} \times \frac{{{\pi ^3}m{e^4}}}{{{h^3}}}{K^2}$$