QBManager

Dual nature of radiation and Matter

142619 An element with atomic number $Z = 11$ emits $E_{α}$ - X-ray of wavelength $λ$ . The atomic number which emits $E_{α} - X$ -ray of wavelength $4 λ$ is

1 4

2 6

3 11

4 44

Explanation:

B Given,
Atomic $Number (Z_{1}) = 1$
Wavelength $(λ_{1}) = λ$
Wavelength of second element $(λ_{2}) = 4 λ$ .
Then atomic number of second element $(Z_{2}) =$ ?
From Moseley's law-
$\sqrt{f} = a (Z - b)$
$f = a^{2} (Z - b)^{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{(Z_{2} - 1)}{(Z_{1} - 1)} (∵ f = \frac{1}{λ})$
$\frac{λ}{4 λ} = \frac{{(Z_{2} - 1)}^{2}}{(11 - 1)^{2}}$
$\frac{100}{4} = {(Z_{2} - 1)}^{2}$
$Z_{2} - 1 = 5$
$Z_{2} = 6$

VITEEE-2013

Dual nature of radiation and Matter

142621 The graph between the square root of the frequency of a specific line of characteristic spectrum of $X$ - ray and the atomic number of the target will be

1

2

3

4

Explanation:

B Relationship between frequency of characteristic spectrum of X-ray and atomic number -
$v = a^{2} (Z - b)^{2}$
Where, $v =$ frequency $Z =$ atomic number
a \& b are constant
$\sqrt{v} = a Z - a b$
We know, straight line equation
$Y = mx + c$
Comparing equation (i) \& (ii), we get graph like this

VITEEE-2012

Dual nature of radiation and Matter

142625 If an X-ray tube is operated at $15 kV$ , then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be

1

7.26 \times 10^{7} m / s, 0.825 \AA

2

3 \times 10^{8} m / s, 1.08 \AA

3

2.7 \times 10^{9} m / s, 2.05 \AA

4 None of the above

Explanation:

A Given,
Operating voltage $(V_{0}) = 15 kV = 15 \times 10^{3} V$
From energy conservation
$\begin{matrix} \frac{1}{2} {mv}^{2} = eV \\ \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2} = 1.6 \times 10^{- 19} \times 15 \times 10^{3} \\ v = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 15 \times 10^{3}}{9.1 \times 10^{- 31}}} \\ = 7.26 \times 10^{7} m / s \end{matrix}$
Lower limit of wavelength produced by X-ray
${eV}_{0} = \frac{hc}{λ_{min}}$
$λ_{min} = \frac{hc}{{eV}_{0}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 15 \times 10^{3}}$
$λ_{min} = 0.825 \times 10^{- 10}$
$λ_{min} = 0.825 \AA$

UP CPMT-2013

Dual nature of radiation and Matter

142626 When a beam of accelerated electrons hits a target, a continuous $X$ -rays spectrum is emitted from the target. Which of the following wavelength is absent in the $X$ -ray spectrum, if the X-ray tube is operating at $40000 V$ ?

1

0.5 \AA

2

1.0 \AA

3

0.25 \AA

4

1.5 \AA

Explanation:

C X-ray tube operating (V) $= 40000 V$
We know,
$λ_{sutoff} = \frac{hc}{eV}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 40000}$
$λ_{sutofi} = 0.31 \AA$
$λ > 0.31 \AA$ wavelength is absent in the $X$ -ray spectrum.

UP CPMT-2011

Dual nature of radiation and Matter

142619 An element with atomic number $Z = 11$ emits $E_{α}$ - X-ray of wavelength $λ$ . The atomic number which emits $E_{α} - X$ -ray of wavelength $4 λ$ is

1 4

2 6

3 11

4 44

Explanation:

B Given,
Atomic $Number (Z_{1}) = 1$
Wavelength $(λ_{1}) = λ$
Wavelength of second element $(λ_{2}) = 4 λ$ .
Then atomic number of second element $(Z_{2}) =$ ?
From Moseley's law-
$\sqrt{f} = a (Z - b)$
$f = a^{2} (Z - b)^{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{(Z_{2} - 1)}{(Z_{1} - 1)} (∵ f = \frac{1}{λ})$
$\frac{λ}{4 λ} = \frac{{(Z_{2} - 1)}^{2}}{(11 - 1)^{2}}$
$\frac{100}{4} = {(Z_{2} - 1)}^{2}$
$Z_{2} - 1 = 5$
$Z_{2} = 6$

VITEEE-2013

Dual nature of radiation and Matter

142621 The graph between the square root of the frequency of a specific line of characteristic spectrum of $X$ - ray and the atomic number of the target will be

1

2

3

4

Explanation:

B Relationship between frequency of characteristic spectrum of X-ray and atomic number -
$v = a^{2} (Z - b)^{2}$
Where, $v =$ frequency $Z =$ atomic number
a \& b are constant
$\sqrt{v} = a Z - a b$
We know, straight line equation
$Y = mx + c$
Comparing equation (i) \& (ii), we get graph like this

VITEEE-2012

Dual nature of radiation and Matter

142625 If an X-ray tube is operated at $15 kV$ , then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be

1

7.26 \times 10^{7} m / s, 0.825 \AA

2

3 \times 10^{8} m / s, 1.08 \AA

3

2.7 \times 10^{9} m / s, 2.05 \AA

4 None of the above

Explanation:

A Given,
Operating voltage $(V_{0}) = 15 kV = 15 \times 10^{3} V$
From energy conservation
$\begin{matrix} \frac{1}{2} {mv}^{2} = eV \\ \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2} = 1.6 \times 10^{- 19} \times 15 \times 10^{3} \\ v = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 15 \times 10^{3}}{9.1 \times 10^{- 31}}} \\ = 7.26 \times 10^{7} m / s \end{matrix}$
Lower limit of wavelength produced by X-ray
${eV}_{0} = \frac{hc}{λ_{min}}$
$λ_{min} = \frac{hc}{{eV}_{0}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 15 \times 10^{3}}$
$λ_{min} = 0.825 \times 10^{- 10}$
$λ_{min} = 0.825 \AA$

UP CPMT-2013

Dual nature of radiation and Matter

142626 When a beam of accelerated electrons hits a target, a continuous $X$ -rays spectrum is emitted from the target. Which of the following wavelength is absent in the $X$ -ray spectrum, if the X-ray tube is operating at $40000 V$ ?

1

0.5 \AA

2

1.0 \AA

3

0.25 \AA

4

1.5 \AA

Explanation:

C X-ray tube operating (V) $= 40000 V$
We know,
$λ_{sutoff} = \frac{hc}{eV}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 40000}$
$λ_{sutofi} = 0.31 \AA$
$λ > 0.31 \AA$ wavelength is absent in the $X$ -ray spectrum.

UP CPMT-2011

Dual nature of radiation and Matter

142619 An element with atomic number $Z = 11$ emits $E_{α}$ - X-ray of wavelength $λ$ . The atomic number which emits $E_{α} - X$ -ray of wavelength $4 λ$ is

1 4

2 6

3 11

4 44

Explanation:

B Given,
Atomic $Number (Z_{1}) = 1$
Wavelength $(λ_{1}) = λ$
Wavelength of second element $(λ_{2}) = 4 λ$ .
Then atomic number of second element $(Z_{2}) =$ ?
From Moseley's law-
$\sqrt{f} = a (Z - b)$
$f = a^{2} (Z - b)^{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{(Z_{2} - 1)}{(Z_{1} - 1)} (∵ f = \frac{1}{λ})$
$\frac{λ}{4 λ} = \frac{{(Z_{2} - 1)}^{2}}{(11 - 1)^{2}}$
$\frac{100}{4} = {(Z_{2} - 1)}^{2}$
$Z_{2} - 1 = 5$
$Z_{2} = 6$

VITEEE-2013

Dual nature of radiation and Matter

142621 The graph between the square root of the frequency of a specific line of characteristic spectrum of $X$ - ray and the atomic number of the target will be

1

2

3

4

Explanation:

B Relationship between frequency of characteristic spectrum of X-ray and atomic number -
$v = a^{2} (Z - b)^{2}$
Where, $v =$ frequency $Z =$ atomic number
a \& b are constant
$\sqrt{v} = a Z - a b$
We know, straight line equation
$Y = mx + c$
Comparing equation (i) \& (ii), we get graph like this

VITEEE-2012

Dual nature of radiation and Matter

142625 If an X-ray tube is operated at $15 kV$ , then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be

1

7.26 \times 10^{7} m / s, 0.825 \AA

2

3 \times 10^{8} m / s, 1.08 \AA

3

2.7 \times 10^{9} m / s, 2.05 \AA

4 None of the above

Explanation:

A Given,
Operating voltage $(V_{0}) = 15 kV = 15 \times 10^{3} V$
From energy conservation
$\begin{matrix} \frac{1}{2} {mv}^{2} = eV \\ \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2} = 1.6 \times 10^{- 19} \times 15 \times 10^{3} \\ v = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 15 \times 10^{3}}{9.1 \times 10^{- 31}}} \\ = 7.26 \times 10^{7} m / s \end{matrix}$
Lower limit of wavelength produced by X-ray
${eV}_{0} = \frac{hc}{λ_{min}}$
$λ_{min} = \frac{hc}{{eV}_{0}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 15 \times 10^{3}}$
$λ_{min} = 0.825 \times 10^{- 10}$
$λ_{min} = 0.825 \AA$

UP CPMT-2013

Dual nature of radiation and Matter

142626 When a beam of accelerated electrons hits a target, a continuous $X$ -rays spectrum is emitted from the target. Which of the following wavelength is absent in the $X$ -ray spectrum, if the X-ray tube is operating at $40000 V$ ?

1

0.5 \AA

2

1.0 \AA

3

0.25 \AA

4

1.5 \AA

Explanation:

C X-ray tube operating (V) $= 40000 V$
We know,
$λ_{sutoff} = \frac{hc}{eV}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 40000}$
$λ_{sutofi} = 0.31 \AA$
$λ > 0.31 \AA$ wavelength is absent in the $X$ -ray spectrum.

UP CPMT-2011

Dual nature of radiation and Matter

142619 An element with atomic number $Z = 11$ emits $E_{α}$ - X-ray of wavelength $λ$ . The atomic number which emits $E_{α} - X$ -ray of wavelength $4 λ$ is

1 4

2 6

3 11

4 44

Explanation:

B Given,
Atomic $Number (Z_{1}) = 1$
Wavelength $(λ_{1}) = λ$
Wavelength of second element $(λ_{2}) = 4 λ$ .
Then atomic number of second element $(Z_{2}) =$ ?
From Moseley's law-
$\sqrt{f} = a (Z - b)$
$f = a^{2} (Z - b)^{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{(Z_{2} - 1)}{(Z_{1} - 1)} (∵ f = \frac{1}{λ})$
$\frac{λ}{4 λ} = \frac{{(Z_{2} - 1)}^{2}}{(11 - 1)^{2}}$
$\frac{100}{4} = {(Z_{2} - 1)}^{2}$
$Z_{2} - 1 = 5$
$Z_{2} = 6$

VITEEE-2013

Dual nature of radiation and Matter

142621 The graph between the square root of the frequency of a specific line of characteristic spectrum of $X$ - ray and the atomic number of the target will be

1

2

3

4

Explanation:

B Relationship between frequency of characteristic spectrum of X-ray and atomic number -
$v = a^{2} (Z - b)^{2}$
Where, $v =$ frequency $Z =$ atomic number
a \& b are constant
$\sqrt{v} = a Z - a b$
We know, straight line equation
$Y = mx + c$
Comparing equation (i) \& (ii), we get graph like this

VITEEE-2012

Dual nature of radiation and Matter

142625 If an X-ray tube is operated at $15 kV$ , then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be

1

7.26 \times 10^{7} m / s, 0.825 \AA

2

3 \times 10^{8} m / s, 1.08 \AA

3

2.7 \times 10^{9} m / s, 2.05 \AA

4 None of the above

Explanation:

A Given,
Operating voltage $(V_{0}) = 15 kV = 15 \times 10^{3} V$
From energy conservation
$\begin{matrix} \frac{1}{2} {mv}^{2} = eV \\ \frac{1}{2} \times 9.1 \times 10^{- 31} \times v^{2} = 1.6 \times 10^{- 19} \times 15 \times 10^{3} \\ v = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 15 \times 10^{3}}{9.1 \times 10^{- 31}}} \\ = 7.26 \times 10^{7} m / s \end{matrix}$
Lower limit of wavelength produced by X-ray
${eV}_{0} = \frac{hc}{λ_{min}}$
$λ_{min} = \frac{hc}{{eV}_{0}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 15 \times 10^{3}}$
$λ_{min} = 0.825 \times 10^{- 10}$
$λ_{min} = 0.825 \AA$

UP CPMT-2013

Dual nature of radiation and Matter

142626 When a beam of accelerated electrons hits a target, a continuous $X$ -rays spectrum is emitted from the target. Which of the following wavelength is absent in the $X$ -ray spectrum, if the X-ray tube is operating at $40000 V$ ?

1

0.5 \AA

2

1.0 \AA

3

0.25 \AA

4

1.5 \AA

Explanation:

C X-ray tube operating (V) $= 40000 V$
We know,
$λ_{sutoff} = \frac{hc}{eV}$
$= \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 40000}$
$λ_{sutofi} = 0.31 \AA$
$λ > 0.31 \AA$ wavelength is absent in the $X$ -ray spectrum.

UP CPMT-2011