Question

plzzzzz....... help !!!? ​

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 :

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

$\rule{300}{1}$

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,

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

$\rule{300}{1}$

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

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

$\rule{300}{1}$

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

$\rule{300}{1}$

$\rule{300}{1}$

Answer 2

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

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 :

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

$\rule{300}{1}$

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,

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

$\rule{300}{1}$

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

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

$\rule{300}{1}$

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

$\rule{300}{1}$

$\rule{300}{1}$