Question

calculate the momentum of a partical which has de Broglic wavelength of 2A°[h=6.626×10*-34 kg m*2 s*-1​

Answer 1

ANSWER:

• The momentum of the particle = 3.313 × 10⁻²⁴ kgm/s.

GIVEN:

• De Broglie wavelength is 2 A°.

TO FIND:

• The momentum of the particle.

EXPLANATION:

${\huge{ \star}} \ \boxed{ \boldsymbol{ \large{ \pink {p =\dfrac{h }{ \lambda}}}}}$

$\sf \implies h = 6.626 \times {10}^{ - 34} \ Js$

$\sf \implies\lambda = 2\ A^{\circ} = 2\times 10^{-10}\ m$

$\sf \leadsto p =\dfrac{6.626 \times {10}^{ - 34} }{2 \times {10}^{ - 10} }$

$\sf \leadsto p =\dfrac{3.313 \times {10}^{ - 34} }{ {10}^{ - 10} }$

$\sf \leadsto p =3.313 \times {10}^{ - 24} \ kgm {s}^{ - 1}$

Hence the momentum of the particle = 3.313 × 10⁻²⁴ kgm/s.

EXTRA CACULATION:

${\huge{ \star }}\ \boxed{ \boldsymbol{ \large{ \orange{E=\dfrac{hc }{ \lambda}}}}}$

$\sf \implies h = 6.626 \times {10}^{ - 34} \ Js$

$\sf \implies\lambda = 2\ A^{\circ} = 2\times 10^{-10}\ m$

$\sf \implies c = 3 \times {10}^{ 8} \ m {s}^{ - 1}$

$\sf \longmapsto E=\dfrac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{2 \times {10}^{ - 10} }$

$\sf \longmapsto E=\dfrac{3.313 \times {10}^{ - 34} \times 3 \times {10}^{8} }{ {10}^{ - 10} }$

$\sf \longmapsto E=9.939 \times {10}^{ - 24} \times {10}^{8}$

$\sf \longmapsto E=9.939 \times {10}^{ - 16} \ J$

Hence the energy of the particle = 9.939 × 10⁻¹⁶ J.

Answer 2

$\huge{\underline{Solution}}$

Momentum = $\sf\dfrac{λ}{h}$

(using de Broglie equation)

$⟹\sf\dfrac{2.5×10 −10}{6.6×10-34}$

$\sf{⟹\: 2.64×10 −24kg\: m\: sec\:</p><p>−1}$

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