QBManager

Structure of Atom

238744 The de-Broglie wavelength $(λ)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $[v_{0}$ is threshold frequency]

1

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{4}}}

2

λ \propto \frac{1}{{(v - v_{0})}^{\frac{3}{2}}}

3

λ \propto \frac{1}{(v - v_{0})}

4

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}}

Explanation:

: For electron
$λ_{DB} = \frac{λ}{\sqrt{2 mK . E}} (de-broglie wavelength)$
Where $λ =$ Wavelength of particles
$h =$ Planck's constant
K.E = Kinetic energy
According to photoelectric effect
$\begin{aligned} h v = h v_{0} + KE \\ KE = h v - h v_{0} \\ λ_{DB} = \frac{h}{\sqrt{2 m \times (h v - h v_{0})}}, λ_{DB} \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}} \end{aligned}$

[JEE Main 2019

Structure of Atom

238746 The wavelength of a ball of mass $100 g$ moving with a velocity of $100 {ms}^{- 1}$ be

1

6.626 \times 10^{- 30} m

2

6.626 \times 10^{- 35} m

3

6.626 \times 10^{- 32} m

4

6.626 \times 10^{- 34} m

Explanation:

: Given that, $m = 100 g, v = 100 m / s$
According to the de-Broglie equation -
$\begin{aligned} λ = \frac{h}{mv} & = \frac{6.626 \times 10^{- 34} kg - m^{2} / \sec}{0.1 kg \times 100 m / \sec} \\ = 6.626 \times 10^{- 35} m \end{aligned}$

Assam CEE-2020

Structure of Atom

238747 The energy ratio of a photon of wavelength $3000 \AA$ and $6000 \AA$ is

1

1 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

: Given that, $λ_{1} = 3000 \AA, λ_{2} = 6000 \AA$
$\begin{aligned} E_{1} = \frac{hc}{λ_{1}} = \frac{hc}{3000} and E_{2} = \frac{hc}{λ_{2}} = \frac{hc}{6000} \\ \frac{E_{1}}{E_{2}} = \frac{\frac{hc}{3000}}{\frac{hc}{6000}} = \frac{hc}{3000} \times \frac{6000}{hc} = \frac{2}{1} \\ E_{1} : E_{2} = 2 : 1 \end{aligned}$

BCECE-2007

Structure of Atom

238748 A photon having a wavelength of $845 \AA$ , causes the ionisation of $N$ atom. What is the ionisation energy of $N$ ?

1

1.4 kJ

2

1.4 \times 10^{4} kJ

3

1.4 \times 10^{2} kJ

4

1.4 \times 10^{3} kJ

Explanation:

: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom.
$∴$ Ionisation energy of nitrogen
$=$ energy of photon
$= Nh \frac{c}{λ}$
Where
$\begin{aligned} N = 6.02 \times 10^{23} \\ c = 3 \times 10^{8} \\ λ = 854 \AA = 854 \times 10^{- 10} M \\ = \frac{6.02 \times 10^{23} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{854 \times 10^{- 10}} \\ = 1.4 \times 10^{6} J {mol}^{- 1} = 1.4 \times 10^{3} kJ {mol}^{- 1} \end{aligned}$

BCECE-2003

Structure of Atom

238749 The increasing order of wavelength for ${He}^{+}$ ion, neutron (n) and electron (e) particles, moving with the same velocity is-

1

λ_{{He}^{+}} < λ_{e} < λ_{n}

2

λ_{{He}^{+}} = λ_{n} = λ_{e}

3

λ_{{He}^{+}} < λ_{n} < λ_{e}

4

λ_{e} < λ_{n} < λ_{{He}^{+}}

Explanation:

$: λ = \frac{h}{m \cdot v}$
(As per de-Broglie relation)
$\begin{aligned} ∴ λ \propto \frac{1}{m} (h and v \\ Also; ∵ m_{{He}^{+}} > m_{n} > m_{e} \\ ∴ λ_{{He}^{+}} < λ_{n} < λ_{e} \end{aligned}$

BCECE-2016

Structure of Atom

238744 The de-Broglie wavelength $(λ)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $[v_{0}$ is threshold frequency]

1

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{4}}}

2

λ \propto \frac{1}{{(v - v_{0})}^{\frac{3}{2}}}

3

λ \propto \frac{1}{(v - v_{0})}

4

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}}

Explanation:

: For electron
$λ_{DB} = \frac{λ}{\sqrt{2 mK . E}} (de-broglie wavelength)$
Where $λ =$ Wavelength of particles
$h =$ Planck's constant
K.E = Kinetic energy
According to photoelectric effect
$\begin{aligned} h v = h v_{0} + KE \\ KE = h v - h v_{0} \\ λ_{DB} = \frac{h}{\sqrt{2 m \times (h v - h v_{0})}}, λ_{DB} \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}} \end{aligned}$

[JEE Main 2019

Structure of Atom

238746 The wavelength of a ball of mass $100 g$ moving with a velocity of $100 {ms}^{- 1}$ be

1

6.626 \times 10^{- 30} m

2

6.626 \times 10^{- 35} m

3

6.626 \times 10^{- 32} m

4

6.626 \times 10^{- 34} m

Explanation:

: Given that, $m = 100 g, v = 100 m / s$
According to the de-Broglie equation -
$\begin{aligned} λ = \frac{h}{mv} & = \frac{6.626 \times 10^{- 34} kg - m^{2} / \sec}{0.1 kg \times 100 m / \sec} \\ = 6.626 \times 10^{- 35} m \end{aligned}$

Assam CEE-2020

Structure of Atom

238747 The energy ratio of a photon of wavelength $3000 \AA$ and $6000 \AA$ is

1

1 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

: Given that, $λ_{1} = 3000 \AA, λ_{2} = 6000 \AA$
$\begin{aligned} E_{1} = \frac{hc}{λ_{1}} = \frac{hc}{3000} and E_{2} = \frac{hc}{λ_{2}} = \frac{hc}{6000} \\ \frac{E_{1}}{E_{2}} = \frac{\frac{hc}{3000}}{\frac{hc}{6000}} = \frac{hc}{3000} \times \frac{6000}{hc} = \frac{2}{1} \\ E_{1} : E_{2} = 2 : 1 \end{aligned}$

BCECE-2007

Structure of Atom

238748 A photon having a wavelength of $845 \AA$ , causes the ionisation of $N$ atom. What is the ionisation energy of $N$ ?

1

1.4 kJ

2

1.4 \times 10^{4} kJ

3

1.4 \times 10^{2} kJ

4

1.4 \times 10^{3} kJ

Explanation:

: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom.
$∴$ Ionisation energy of nitrogen
$=$ energy of photon
$= Nh \frac{c}{λ}$
Where
$\begin{aligned} N = 6.02 \times 10^{23} \\ c = 3 \times 10^{8} \\ λ = 854 \AA = 854 \times 10^{- 10} M \\ = \frac{6.02 \times 10^{23} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{854 \times 10^{- 10}} \\ = 1.4 \times 10^{6} J {mol}^{- 1} = 1.4 \times 10^{3} kJ {mol}^{- 1} \end{aligned}$

BCECE-2003

Structure of Atom

238749 The increasing order of wavelength for ${He}^{+}$ ion, neutron (n) and electron (e) particles, moving with the same velocity is-

1

λ_{{He}^{+}} < λ_{e} < λ_{n}

2

λ_{{He}^{+}} = λ_{n} = λ_{e}

3

λ_{{He}^{+}} < λ_{n} < λ_{e}

4

λ_{e} < λ_{n} < λ_{{He}^{+}}

Explanation:

$: λ = \frac{h}{m \cdot v}$
(As per de-Broglie relation)
$\begin{aligned} ∴ λ \propto \frac{1}{m} (h and v \\ Also; ∵ m_{{He}^{+}} > m_{n} > m_{e} \\ ∴ λ_{{He}^{+}} < λ_{n} < λ_{e} \end{aligned}$

BCECE-2016

Structure of Atom

238744 The de-Broglie wavelength $(λ)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $[v_{0}$ is threshold frequency]

1

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{4}}}

2

λ \propto \frac{1}{{(v - v_{0})}^{\frac{3}{2}}}

3

λ \propto \frac{1}{(v - v_{0})}

4

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}}

Explanation:

: For electron
$λ_{DB} = \frac{λ}{\sqrt{2 mK . E}} (de-broglie wavelength)$
Where $λ =$ Wavelength of particles
$h =$ Planck's constant
K.E = Kinetic energy
According to photoelectric effect
$\begin{aligned} h v = h v_{0} + KE \\ KE = h v - h v_{0} \\ λ_{DB} = \frac{h}{\sqrt{2 m \times (h v - h v_{0})}}, λ_{DB} \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}} \end{aligned}$

[JEE Main 2019

Structure of Atom

238746 The wavelength of a ball of mass $100 g$ moving with a velocity of $100 {ms}^{- 1}$ be

1

6.626 \times 10^{- 30} m

2

6.626 \times 10^{- 35} m

3

6.626 \times 10^{- 32} m

4

6.626 \times 10^{- 34} m

Explanation:

: Given that, $m = 100 g, v = 100 m / s$
According to the de-Broglie equation -
$\begin{aligned} λ = \frac{h}{mv} & = \frac{6.626 \times 10^{- 34} kg - m^{2} / \sec}{0.1 kg \times 100 m / \sec} \\ = 6.626 \times 10^{- 35} m \end{aligned}$

Assam CEE-2020

Structure of Atom

238747 The energy ratio of a photon of wavelength $3000 \AA$ and $6000 \AA$ is

1

1 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

: Given that, $λ_{1} = 3000 \AA, λ_{2} = 6000 \AA$
$\begin{aligned} E_{1} = \frac{hc}{λ_{1}} = \frac{hc}{3000} and E_{2} = \frac{hc}{λ_{2}} = \frac{hc}{6000} \\ \frac{E_{1}}{E_{2}} = \frac{\frac{hc}{3000}}{\frac{hc}{6000}} = \frac{hc}{3000} \times \frac{6000}{hc} = \frac{2}{1} \\ E_{1} : E_{2} = 2 : 1 \end{aligned}$

BCECE-2007

Structure of Atom

238748 A photon having a wavelength of $845 \AA$ , causes the ionisation of $N$ atom. What is the ionisation energy of $N$ ?

1

1.4 kJ

2

1.4 \times 10^{4} kJ

3

1.4 \times 10^{2} kJ

4

1.4 \times 10^{3} kJ

Explanation:

: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom.
$∴$ Ionisation energy of nitrogen
$=$ energy of photon
$= Nh \frac{c}{λ}$
Where
$\begin{aligned} N = 6.02 \times 10^{23} \\ c = 3 \times 10^{8} \\ λ = 854 \AA = 854 \times 10^{- 10} M \\ = \frac{6.02 \times 10^{23} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{854 \times 10^{- 10}} \\ = 1.4 \times 10^{6} J {mol}^{- 1} = 1.4 \times 10^{3} kJ {mol}^{- 1} \end{aligned}$

BCECE-2003

Structure of Atom

238749 The increasing order of wavelength for ${He}^{+}$ ion, neutron (n) and electron (e) particles, moving with the same velocity is-

1

λ_{{He}^{+}} < λ_{e} < λ_{n}

2

λ_{{He}^{+}} = λ_{n} = λ_{e}

3

λ_{{He}^{+}} < λ_{n} < λ_{e}

4

λ_{e} < λ_{n} < λ_{{He}^{+}}

Explanation:

$: λ = \frac{h}{m \cdot v}$
(As per de-Broglie relation)
$\begin{aligned} ∴ λ \propto \frac{1}{m} (h and v \\ Also; ∵ m_{{He}^{+}} > m_{n} > m_{e} \\ ∴ λ_{{He}^{+}} < λ_{n} < λ_{e} \end{aligned}$

BCECE-2016

Structure of Atom

238744 The de-Broglie wavelength $(λ)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $[v_{0}$ is threshold frequency]

1

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{4}}}

2

λ \propto \frac{1}{{(v - v_{0})}^{\frac{3}{2}}}

3

λ \propto \frac{1}{(v - v_{0})}

4

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}}

Explanation:

: For electron
$λ_{DB} = \frac{λ}{\sqrt{2 mK . E}} (de-broglie wavelength)$
Where $λ =$ Wavelength of particles
$h =$ Planck's constant
K.E = Kinetic energy
According to photoelectric effect
$\begin{aligned} h v = h v_{0} + KE \\ KE = h v - h v_{0} \\ λ_{DB} = \frac{h}{\sqrt{2 m \times (h v - h v_{0})}}, λ_{DB} \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}} \end{aligned}$

[JEE Main 2019

Structure of Atom

238746 The wavelength of a ball of mass $100 g$ moving with a velocity of $100 {ms}^{- 1}$ be

1

6.626 \times 10^{- 30} m

2

6.626 \times 10^{- 35} m

3

6.626 \times 10^{- 32} m

4

6.626 \times 10^{- 34} m

Explanation:

: Given that, $m = 100 g, v = 100 m / s$
According to the de-Broglie equation -
$\begin{aligned} λ = \frac{h}{mv} & = \frac{6.626 \times 10^{- 34} kg - m^{2} / \sec}{0.1 kg \times 100 m / \sec} \\ = 6.626 \times 10^{- 35} m \end{aligned}$

Assam CEE-2020

Structure of Atom

238747 The energy ratio of a photon of wavelength $3000 \AA$ and $6000 \AA$ is

1

1 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

: Given that, $λ_{1} = 3000 \AA, λ_{2} = 6000 \AA$
$\begin{aligned} E_{1} = \frac{hc}{λ_{1}} = \frac{hc}{3000} and E_{2} = \frac{hc}{λ_{2}} = \frac{hc}{6000} \\ \frac{E_{1}}{E_{2}} = \frac{\frac{hc}{3000}}{\frac{hc}{6000}} = \frac{hc}{3000} \times \frac{6000}{hc} = \frac{2}{1} \\ E_{1} : E_{2} = 2 : 1 \end{aligned}$

BCECE-2007

Structure of Atom

238748 A photon having a wavelength of $845 \AA$ , causes the ionisation of $N$ atom. What is the ionisation energy of $N$ ?

1

1.4 kJ

2

1.4 \times 10^{4} kJ

3

1.4 \times 10^{2} kJ

4

1.4 \times 10^{3} kJ

Explanation:

: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom.
$∴$ Ionisation energy of nitrogen
$=$ energy of photon
$= Nh \frac{c}{λ}$
Where
$\begin{aligned} N = 6.02 \times 10^{23} \\ c = 3 \times 10^{8} \\ λ = 854 \AA = 854 \times 10^{- 10} M \\ = \frac{6.02 \times 10^{23} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{854 \times 10^{- 10}} \\ = 1.4 \times 10^{6} J {mol}^{- 1} = 1.4 \times 10^{3} kJ {mol}^{- 1} \end{aligned}$

BCECE-2003

Structure of Atom

238749 The increasing order of wavelength for ${He}^{+}$ ion, neutron (n) and electron (e) particles, moving with the same velocity is-

1

λ_{{He}^{+}} < λ_{e} < λ_{n}

2

λ_{{He}^{+}} = λ_{n} = λ_{e}

3

λ_{{He}^{+}} < λ_{n} < λ_{e}

4

λ_{e} < λ_{n} < λ_{{He}^{+}}

Explanation:

$: λ = \frac{h}{m \cdot v}$
(As per de-Broglie relation)
$\begin{aligned} ∴ λ \propto \frac{1}{m} (h and v \\ Also; ∵ m_{{He}^{+}} > m_{n} > m_{e} \\ ∴ λ_{{He}^{+}} < λ_{n} < λ_{e} \end{aligned}$

BCECE-2016

Structure of Atom

238744 The de-Broglie wavelength $(λ)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $[v_{0}$ is threshold frequency]

1

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{4}}}

2

λ \propto \frac{1}{{(v - v_{0})}^{\frac{3}{2}}}

3

λ \propto \frac{1}{(v - v_{0})}

4

λ \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}}

Explanation:

: For electron
$λ_{DB} = \frac{λ}{\sqrt{2 mK . E}} (de-broglie wavelength)$
Where $λ =$ Wavelength of particles
$h =$ Planck's constant
K.E = Kinetic energy
According to photoelectric effect
$\begin{aligned} h v = h v_{0} + KE \\ KE = h v - h v_{0} \\ λ_{DB} = \frac{h}{\sqrt{2 m \times (h v - h v_{0})}}, λ_{DB} \propto \frac{1}{{(v - v_{0})}^{\frac{1}{2}}} \end{aligned}$

[JEE Main 2019

Structure of Atom

238746 The wavelength of a ball of mass $100 g$ moving with a velocity of $100 {ms}^{- 1}$ be

1

6.626 \times 10^{- 30} m

2

6.626 \times 10^{- 35} m

3

6.626 \times 10^{- 32} m

4

6.626 \times 10^{- 34} m

Explanation:

: Given that, $m = 100 g, v = 100 m / s$
According to the de-Broglie equation -
$\begin{aligned} λ = \frac{h}{mv} & = \frac{6.626 \times 10^{- 34} kg - m^{2} / \sec}{0.1 kg \times 100 m / \sec} \\ = 6.626 \times 10^{- 35} m \end{aligned}$

Assam CEE-2020

Structure of Atom

238747 The energy ratio of a photon of wavelength $3000 \AA$ and $6000 \AA$ is

1

1 : 1

2

2 : 1

3

1 : 2

4

1 : 4

Explanation:

: Given that, $λ_{1} = 3000 \AA, λ_{2} = 6000 \AA$
$\begin{aligned} E_{1} = \frac{hc}{λ_{1}} = \frac{hc}{3000} and E_{2} = \frac{hc}{λ_{2}} = \frac{hc}{6000} \\ \frac{E_{1}}{E_{2}} = \frac{\frac{hc}{3000}}{\frac{hc}{6000}} = \frac{hc}{3000} \times \frac{6000}{hc} = \frac{2}{1} \\ E_{1} : E_{2} = 2 : 1 \end{aligned}$

BCECE-2007

Structure of Atom

238748 A photon having a wavelength of $845 \AA$ , causes the ionisation of $N$ atom. What is the ionisation energy of $N$ ?

1

1.4 kJ

2

1.4 \times 10^{4} kJ

3

1.4 \times 10^{2} kJ

4

1.4 \times 10^{3} kJ

Explanation:

: The ionisation energy is the amount of energy required to take out most loosely bonded electron from isolated gaseous atom.
$∴$ Ionisation energy of nitrogen
$=$ energy of photon
$= Nh \frac{c}{λ}$
Where
$\begin{aligned} N = 6.02 \times 10^{23} \\ c = 3 \times 10^{8} \\ λ = 854 \AA = 854 \times 10^{- 10} M \\ = \frac{6.02 \times 10^{23} \times 6.6 \times 10^{- 34} \times 3 \times 10^{8}}{854 \times 10^{- 10}} \\ = 1.4 \times 10^{6} J {mol}^{- 1} = 1.4 \times 10^{3} kJ {mol}^{- 1} \end{aligned}$

BCECE-2003

Structure of Atom

238749 The increasing order of wavelength for ${He}^{+}$ ion, neutron (n) and electron (e) particles, moving with the same velocity is-

1

λ_{{He}^{+}} < λ_{e} < λ_{n}

2

λ_{{He}^{+}} = λ_{n} = λ_{e}

3

λ_{{He}^{+}} < λ_{n} < λ_{e}

4

λ_{e} < λ_{n} < λ_{{He}^{+}}

Explanation:

$: λ = \frac{h}{m \cdot v}$
(As per de-Broglie relation)
$\begin{aligned} ∴ λ \propto \frac{1}{m} (h and v \\ Also; ∵ m_{{He}^{+}} > m_{n} > m_{e} \\ ∴ λ_{{He}^{+}} < λ_{n} < λ_{e} \end{aligned}$

BCECE-2016