Question

8. The energy that should be added to an
electron to reduce its de-Broglie wavelength
from 1 nm to 0.5 nm is
(a) four-times the initial energy
(b) equal to the initial energy
(c) two-times the initial energy
(d) three times the initial energy​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Question \colon}}}}$

The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is

(a) four-times the initial energy ✓

(b) equal to the initial energy

(c) two-times the initial energy

(d) three times the initial energy

$\Huge{\underline{\underline{\mathfrak{Solution \colon}}}}$

Option (A) is correct

From the Question,

• Initial Wavelength ($\tt \lambda_1$) = 1 nm

• New Wavelength ($\tt \lambda_2$) = 0.5 nm

Metre - NanoMetre Conversion :

$\boxed{ \boxed{ \tt \: 1 \: nm = {10}^{ - 9} m}}$

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We know that,

$\tt \lambda = \dfrac{h}{p}$

Also,

$\tt \: K = \dfrac{ 1}{2}mv^2 \\ \\ \dashrightarrow \: \tt \: 2K = m \times \dfrac{p^2}{m^2} \\ \\ \dashrightarrow \tt P = \sqrt{2mK}$

Implies,

$\longrightarrow \tt \lambda = h \times \dfrac{1}{p} \\ \\ \longrightarrow \tt \lambda = h \times \dfrac{1}{\sqrt{2mK}} \\ \\ \huge{\longrightarrow \boxed{\boxed{\tt \sqrt{E} \propto \dfrac{1}{\lambda} }}}$

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$\fbox{\sf Know \ The \ Terms }$

$\sf{Here}\begin{cases}\sf{E \longrightarrow Energy}\\\sf{P \longrightarrow Momentum }\\ \sf{ \lambda \longrightarrow Wavelength} \\ \sf{m \longrightarrow Mass } \end{cases}$

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Let the energy that has to be supplied to the particle be n times it's Initial Energy

Thus,

$\tt \sqrt{\dfrac{E_2}{E_1} } = \dfrac{\lambda_1}{\lambda_2}\\ \\ \longrightarrow \tt \ \dfrac{\sqrt{n \cancel{E}}}{\sqrt{\cancel{E}}} = \dfrac{1 \times 10^{-9}}{0.5 \times 10^{-9}} \\ \\ \longrightarrow \tt n = {(\dfrac{1}{0.5})}^{2} \\ \\ \huge{\longrightarrow \boxed{\boxed{\tt n = 4 }}}$

The energy which has to be supplied to the particle is 4 times the initial energy

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Answer 2

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❏ Question:-

@The energy that should be added to an

The energy that should be added to anelectron to reduce its de-Broglie wavelength wavelengthfrom 1 nm to 0.5 nm is ,

(a) four-times the initial energy

(b) equal to the initial energy

(c) two-times the initial energy

(d) three times the initial energy

❏ Solution:-

✏ Given:-

• reducing in wavelength = 1 nm to 0.5 nm

$\blacksquare\:\:\:\sf{\ \ {Initial\: Wavelength=\lambda_1=1\: nm}}$

$\blacksquare\:\:\:\sf{\ \ {final\: Wavelength=\lambda_2=0.5\: nm}}$

✏ To Find:-

• required Energy to reduce that wavelength from 1 nm to 0.5 nm .

✏ Explanation :-

We know that ,

$\blacksquare\:\:\:\boxed{\sf{\ \ {E=h\nu}}}$ -------------(i)

And

$\blacksquare\:\:\:\boxed{\sf{\ \ {C=\nu\lambda}}}$ -------------(ii)

Now from $equ^n(ii)$ we get ,

$\longrightarrow\sf{\ \ {\nu=\dfrac{C}{\lambda}}}$

Now from $equ^n(i)$

$\implies\sf{\ \ {E=h\nu}}$

$\implies\sf{\ \ {E=h\times\dfrac{C}{\lambda}}}$

$\implies\sf{\ \ {E=Ch\times\dfrac{1}{\lambda}}}$

$\implies\boxed{\sf{\ \ {E\infty\dfrac{1}{\lambda}}}}$

therefore ,

$\implies\sf{\ \ {E_1\:\infty\:\dfrac{1}{\lambda_1}}}$

and

$\implies\sf{\ \ {E_2\:\infty\:\dfrac{1}{\lambda_2}}}$

Therefore ,

$\implies\sf{\ \ {\dfrac{E_2}{E_1}=\dfrac{\lambda_1}{\lambda_2}}}$

$\implies\sf{\ \ {\dfrac{E_2}{E_1}=\dfrac{1}{0.5}}}$

$\implies\sf{\ \ {\dfrac{E_2}{E_1}=2}}$

$\implies\large{\boxed{\sf{\ \ {E_2=2E_1}}}}$

Hence , The energy that should be added to an

electron to reduce its de-Broglie wavelength

from 1 nm to 0.5 nm is twice to the initial energy.

➩ option (c) two-times the initial energy

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