Question

solve fastttt it ??!!!!???!! ​

Answer 1

Question :–

$\\ { \bold{show \: \: that \: : \: \lambda = \frac{h}{ \sqrt{3mKT} } }}$

ANSWER :–

• We know that –

$\\ \: \: \: { \huge{.}} \: \: { \bold{E = \dfrac{hc}{ \lambda} }}$

• And Einstein's formula –

$\\ \: \: \: { \huge{.}} \: \: { \bold{E = m {c}^{2} }}$

• From both equation –

$\\ \implies { \bold{\dfrac{hc}{ \lambda} = m {c}^{2} }}$

$\\ \implies { \bold{\dfrac{h}{ \lambda} = m c }}$

$\\ \implies { \bold{\dfrac{h}{ mc} = \lambda }}$

$\\ \implies { \bold{ \lambda = \dfrac{h}{ p} \: \: \: \: \: \: \: \: \: \left[ \because \: \: p = mc \right] \: \: \: \: }}$

• We know that –

$\\ \implies { \bold{kinetic \: \: energy(E) = \frac{1}{2} m {v}^{2} }}$

$\\ \implies { \bold{E = \dfrac{1}{2} \dfrac{ m {}^{2} {v}^{2}}{m} }}$

• We should write this as –

$\\ \implies { \bold{E = \dfrac{1}{2} \dfrac{ {p}^{2} }{m} }}$

$\\ \implies { \bold{ p = \sqrt{2mE} }}$

• So that –

$\\ \implies { \bold{ \lambda = \dfrac{h}{ \sqrt{2mE} } }}$

$\\ \implies { \bold{ \lambda = \dfrac{h}{ \sqrt{2m \left(\dfrac{3}{2}KT \right)} } \: \: \: \: \: \: \: \: \: \left[ \: \because \: \: E = \dfrac{3}{2}KT \right] \: \: \: \:}}$

$\\ \implies { \bold{ \lambda = \dfrac{h}{ \sqrt{3m KT} } }}$

$\\ { \bold{ \underline{ \underline{Used \: \: symbols}} : - }}$

$\\ { \bold{\: \: { \huge{.}} \: \lambda =wavelength}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: c =speed \: \: of \: \: light}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: h=Planck's \: \: constant}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: E=energy}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: m=mass \: \: of \: \: partical}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: P=Linear \: \: momentum}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: K=Boltzman's \: \: constant}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: T=Temperature}}$

Answer 2

Question :–

$\\ { \bold{show \: \: that \: : \: \lambda = \frac{h}{ \sqrt{3mKT} } }}$

ANSWER :–

• We know that –

$\\ \: \: \: { \huge{.}} \: \: { \bold{E = \dfrac{hc}{ \lambda} }}$

• And Einstein's formula –

$\\ \: \: \: { \huge{.}} \: \: { \bold{E = m {c}^{2} }}$

• From both equation –

$\\ \implies { \bold{\dfrac{hc}{ \lambda} = m {c}^{2} }}$

$\\ \implies { \bold{\dfrac{h}{ \lambda} = m c }}$

$\\ \implies { \bold{\dfrac{h}{ mc} = \lambda }}$

$\\ \implies { \bold{ \lambda = \dfrac{h}{ p} \: \: \: \: \: \: \: \: \: \left[ \because \: \: p = mc \right] \: \: \: \: }}$

• We know that –

$\\ \implies { \bold{kinetic \: \: energy(E) = \frac{1}{2} m {v}^{2} }}$

$\\ \implies { \bold{E = \dfrac{1}{2} \dfrac{ m {}^{2} {v}^{2}}{m} }}$

• We should write this as –

$\\ \implies { \bold{E = \dfrac{1}{2} \dfrac{ {p}^{2} }{m} }}$

$\\ \implies { \bold{ p = \sqrt{2mE} }}$

• So that –

$\\ \implies { \bold{ \lambda = \dfrac{h}{ \sqrt{2mE} } }}$

$\\ \implies { \bold{ \lambda = \dfrac{h}{ \sqrt{2m \left(\dfrac{3}{2}KT \right)} } \: \: \: \: \: \: \: \: \: \left[ \: \because \: \: E = \dfrac{3}{2}KT \right] \: \: \: \:}}$

$\\ \implies { \bold{ \lambda = \dfrac{h}{ \sqrt{3m KT} } }}$

$\\ { \bold{ \underline{ \underline{Used \: \: symbols}} : - }}$

$\\ { \bold{\: \: { \huge{.}} \: \lambda =wavelength}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: c =speed \: \: of \: \: light}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: h=Planck's \: \: constant}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: E=energy}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: m=mass \: \: of \: \: partical}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: P=Linear \: \: momentum}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: K=Boltzman's \: \: constant}}$

$\\ \: \: { \huge{.}} { \bold{ \: \: T=Temperature}}$