An electron is accelerated through a potential difference of $10, 000 V$ . Its de Broglie wavelength is (nearly) $(m_{e} = 9 \times 10^{- 31} k g)$

Question

An electron is accelerated through a potential difference of 10-000 V- Its de Broglie wavelength is- -nearly- - -m-e - 9 times 10-31- kg-

Toppr Admin · Accepted Answer

Correct option is B- -12-2 -times 10-12- m-For an electron accelerated through a potential V-lambda - -dfrac-12-27-sqrt-V-160-160-mathring - A - - -dfrac-12-27 -times 10-10-sqrt-10000- - 12-27 -times 10-12- m-

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) :
$$(m_e = 9 \times 10^{-31} kg)$$

Correct option is B. $$12.2 \times 10^{-12} m$$
For an electron accelerated through a potential V,

$$\lambda = \dfrac{12.27}{\sqrt{V}} \mathring { A } $$

$$= \dfrac{12.27 \times 10^{-10}}{\sqrt{10000}}$$

$$ = 12.27 \times 10^{-12} m$$

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : $$(m_e = 9 \times 10^{-31} kg)$$

Correct option is B. $$12.2 \times 10^{-12} m$$For an electron accelerated through a potential V,$$\lambda = \dfrac{12.27}{\sqrt{V}} \mathring { A } $$$$= \dfrac{12.27 \times 10^{-10}}{\sqrt{10000}}$$$$ = 12.27 \times 10^{-12} m$$

An electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) :
$$(m_e = 9 \times 10^{-31} kg)$$

Correct option is B. $$12.2 \times 10^{-12} m$$
For an electron accelerated through a potential V,

$$\lambda = \dfrac{12.27}{\sqrt{V}} \mathring { A } $$

$$= \dfrac{12.27 \times 10^{-10}}{\sqrt{10000}}$$

$$ = 12.27 \times 10^{-12} m$$