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TEST SERIES (PHYSICS FST)

266260 If the momentum of electron is changes by $P$ , then the de-Broglie wavelength associated with it changes by $0.5 %$ . The initial momentum of electron will be:

1 250 P

2 200 P

3

\frac{P}{200}

4 100 P

Explanation:

b
$\begin{aligned} λ = \frac{h}{P} \\ \frac{d λ}{λ} = - \frac{d P}{P} \\ \frac{0.5}{100} = \frac{P}{P^{'}} \\ P^{'} = 200 P \end{aligned}$

**NCERT-XIIIII-285**

TEST SERIES (PHYSICS FST)

266261 A car moves with uniformacceleration up to some distance. Initial and final velocities are $u$ and $v$ then velocity at half way of the car will be:

1

\frac{1}{2} (u + v)

2

\frac{1}{2} (u^{2} + v^{2})

3

\sqrt{\frac{1}{2} (u^{2} + v^{2})}

4

\frac{1}{2} \sqrt{(u^{2} + v^{2})}

Explanation:

c
Suppose velocity at mid point is V .

$\begin{aligned} y^{2} = u^{2} + 2 a s \\ y^{2} = y^{2} + 2 a s \\ y^{2} - v^{2} = u^{2} - v^{2} \\ 2 v^{2} = u^{2} + v^{2} \\ y^{2} = \frac{1}{2} (u^{2} + v^{2}) \\ V = \sqrt{\frac{1}{2} (u^{2} + v^{2})} \end{aligned}$

**NCERT-XI-I-18**

TEST SERIES (PHYSICS FST)

266262 lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):

1 3.4

2 1.89

3 10.2

4 8.5

Explanation:

a
Energy of electron in $n^{th}$ state of hydrogen atom
$E = \frac{- 13.6}{π^{2}} e V$
Energy of electron in second orbit
$E_{2} = \frac{- 13.6}{(2)^{2}} = - 3.4 eV$
Thus 3.4 eVenergyis required to remove an electron from second orbit

**NCERT-XII-II-300**

TEST SERIES (PHYSICS FST)

266263 A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :

1 4000 nm

2 6000 nm

3 4000 A

4 6000 A

Explanation:

c
$A - \frac{12400 eV}{3.1 eV}$ A $= 4000 A$ .

**NTA**

TEST SERIES (PHYSICS FST)

266260 If the momentum of electron is changes by $P$ , then the de-Broglie wavelength associated with it changes by $0.5 %$ . The initial momentum of electron will be:

1 250 P

2 200 P

3

\frac{P}{200}

4 100 P

Explanation:

b
$\begin{aligned} λ = \frac{h}{P} \\ \frac{d λ}{λ} = - \frac{d P}{P} \\ \frac{0.5}{100} = \frac{P}{P^{'}} \\ P^{'} = 200 P \end{aligned}$

**NCERT-XIIIII-285**

TEST SERIES (PHYSICS FST)

266261 A car moves with uniformacceleration up to some distance. Initial and final velocities are $u$ and $v$ then velocity at half way of the car will be:

1

\frac{1}{2} (u + v)

2

\frac{1}{2} (u^{2} + v^{2})

3

\sqrt{\frac{1}{2} (u^{2} + v^{2})}

4

\frac{1}{2} \sqrt{(u^{2} + v^{2})}

Explanation:

c
Suppose velocity at mid point is V .

$\begin{aligned} y^{2} = u^{2} + 2 a s \\ y^{2} = y^{2} + 2 a s \\ y^{2} - v^{2} = u^{2} - v^{2} \\ 2 v^{2} = u^{2} + v^{2} \\ y^{2} = \frac{1}{2} (u^{2} + v^{2}) \\ V = \sqrt{\frac{1}{2} (u^{2} + v^{2})} \end{aligned}$

**NCERT-XI-I-18**

TEST SERIES (PHYSICS FST)

266262 lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):

1 3.4

2 1.89

3 10.2

4 8.5

Explanation:

a
Energy of electron in $n^{th}$ state of hydrogen atom
$E = \frac{- 13.6}{π^{2}} e V$
Energy of electron in second orbit
$E_{2} = \frac{- 13.6}{(2)^{2}} = - 3.4 eV$
Thus 3.4 eVenergyis required to remove an electron from second orbit

**NCERT-XII-II-300**

TEST SERIES (PHYSICS FST)

266263 A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :

1 4000 nm

2 6000 nm

3 4000 A

4 6000 A

Explanation:

c
$A - \frac{12400 eV}{3.1 eV}$ A $= 4000 A$ .

**NTA**

TEST SERIES (PHYSICS FST)

266260 If the momentum of electron is changes by $P$ , then the de-Broglie wavelength associated with it changes by $0.5 %$ . The initial momentum of electron will be:

1 250 P

2 200 P

3

\frac{P}{200}

4 100 P

Explanation:

b
$\begin{aligned} λ = \frac{h}{P} \\ \frac{d λ}{λ} = - \frac{d P}{P} \\ \frac{0.5}{100} = \frac{P}{P^{'}} \\ P^{'} = 200 P \end{aligned}$

**NCERT-XIIIII-285**

TEST SERIES (PHYSICS FST)

266261 A car moves with uniformacceleration up to some distance. Initial and final velocities are $u$ and $v$ then velocity at half way of the car will be:

1

\frac{1}{2} (u + v)

2

\frac{1}{2} (u^{2} + v^{2})

3

\sqrt{\frac{1}{2} (u^{2} + v^{2})}

4

\frac{1}{2} \sqrt{(u^{2} + v^{2})}

Explanation:

c
Suppose velocity at mid point is V .

$\begin{aligned} y^{2} = u^{2} + 2 a s \\ y^{2} = y^{2} + 2 a s \\ y^{2} - v^{2} = u^{2} - v^{2} \\ 2 v^{2} = u^{2} + v^{2} \\ y^{2} = \frac{1}{2} (u^{2} + v^{2}) \\ V = \sqrt{\frac{1}{2} (u^{2} + v^{2})} \end{aligned}$

**NCERT-XI-I-18**

TEST SERIES (PHYSICS FST)

266262 lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):

1 3.4

2 1.89

3 10.2

4 8.5

Explanation:

a
Energy of electron in $n^{th}$ state of hydrogen atom
$E = \frac{- 13.6}{π^{2}} e V$
Energy of electron in second orbit
$E_{2} = \frac{- 13.6}{(2)^{2}} = - 3.4 eV$
Thus 3.4 eVenergyis required to remove an electron from second orbit

**NCERT-XII-II-300**

TEST SERIES (PHYSICS FST)

266263 A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :

1 4000 nm

2 6000 nm

3 4000 A

4 6000 A

Explanation:

c
$A - \frac{12400 eV}{3.1 eV}$ A $= 4000 A$ .

**NTA**

TEST SERIES (PHYSICS FST)

266260 If the momentum of electron is changes by $P$ , then the de-Broglie wavelength associated with it changes by $0.5 %$ . The initial momentum of electron will be:

1 250 P

2 200 P

3

\frac{P}{200}

4 100 P

Explanation:

b
$\begin{aligned} λ = \frac{h}{P} \\ \frac{d λ}{λ} = - \frac{d P}{P} \\ \frac{0.5}{100} = \frac{P}{P^{'}} \\ P^{'} = 200 P \end{aligned}$

**NCERT-XIIIII-285**

TEST SERIES (PHYSICS FST)

266261 A car moves with uniformacceleration up to some distance. Initial and final velocities are $u$ and $v$ then velocity at half way of the car will be:

1

\frac{1}{2} (u + v)

2

\frac{1}{2} (u^{2} + v^{2})

3

\sqrt{\frac{1}{2} (u^{2} + v^{2})}

4

\frac{1}{2} \sqrt{(u^{2} + v^{2})}

Explanation:

c
Suppose velocity at mid point is V .

$\begin{aligned} y^{2} = u^{2} + 2 a s \\ y^{2} = y^{2} + 2 a s \\ y^{2} - v^{2} = u^{2} - v^{2} \\ 2 v^{2} = u^{2} + v^{2} \\ y^{2} = \frac{1}{2} (u^{2} + v^{2}) \\ V = \sqrt{\frac{1}{2} (u^{2} + v^{2})} \end{aligned}$

**NCERT-XI-I-18**

TEST SERIES (PHYSICS FST)

266262 lonisation potential of hydrogen atom is 13.6 eV . The energy required to remove an electron from the second orbit of hydrogen is (in eV ):

1 3.4

2 1.89

3 10.2

4 8.5

Explanation:

a
Energy of electron in $n^{th}$ state of hydrogen atom
$E = \frac{- 13.6}{π^{2}} e V$
Energy of electron in second orbit
$E_{2} = \frac{- 13.6}{(2)^{2}} = - 3.4 eV$
Thus 3.4 eVenergyis required to remove an electron from second orbit

**NCERT-XII-II-300**

TEST SERIES (PHYSICS FST)

266263 A p-n photodiode is fabricated from a semiconductor with a band gap of 3.1 eV . It can detect a signal of wavelength :

1 4000 nm

2 6000 nm

3 4000 A

4 6000 A

Explanation:

c
$A - \frac{12400 eV}{3.1 eV}$ A $= 4000 A$ .

**NTA**

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