An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is (Given me=9.1× 10- 28g,mn=1.674× 10- 24g )

Question

An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is (Given me=9.1× 10- 28g,mn=1.674× 10- 24g ) (A) x (B) (x/1840) (C) 1840 x (D) None

Tardigrade Admin · Accepted Answer

λ electron =λ neutron (h/me Ve)=(h/mn Vn) So Vn=(me Ve/mn)=(9.1 × 1 0- 28 × x/1.674 × 1 0- 24)=(x/1840) .

Q. An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is

(Given $m_{e} = 9.1 \times 1 0^{- 28} g, m_{n} = 1.674 \times 1 0^{- 24} g$ )

x

$\frac{x}{1840}$

1840 x

None

$λ$ electron $= λ$ neutron

$\frac{h}{m _{e} V _{e}} = \frac{h}{m _{n} V _{n}}$

So $Vn = \frac{m _{e} V _{e}}{m _{n}} = \frac{9.1 \times 1 0 ^{- 28} \times x}{1.674 \times 1 0 ^{- 24}} = \frac{x}{1840}$ .

Q. An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is (Given me​=9.1×10−28g,mn​=1.674×10−24g )

x

1840x​

1840 x

None

λ electron =λ neutron me​Ve​h​=mn​Vn​h​ So Vn=mn​me​Ve​​=1.674×10−249.1×10−28×x​=1840x​ .

Q. An electron of a velocity x is found to have a certain value of de Broglie wavelength. The velocity to be possessed by the neutron to have the same de Broglie wavelength is

(Given $m_{e} = 9.1 \times 1 0^{- 28} g, m_{n} = 1.674 \times 1 0^{- 24} g$ )

$\frac{x}{1840}$

$λ$ electron $= λ$ neutron

$\frac{h}{m _{e} V _{e}} = \frac{h}{m _{n} V _{n}}$

So $Vn = \frac{m _{e} V _{e}}{m _{n}} = \frac{9.1 \times 1 0 ^{- 28} \times x}{1.674 \times 1 0 ^{- 24}} = \frac{x}{1840}$ .