The de Broglie wavelength of an electron is 0.4 × 10-10 m when its kinetic energy is 1.0 keV. Its wavelength will be 1.0 × 10-10 m, when its kinetic energy is

Question

The de Broglie wavelength of an electron is 0.4 × 10-10 m when its kinetic energy is 1.0 keV. Its wavelength will be 1.0 × 10-10 m, when its kinetic energy is (A) 0.2 keV (B) 0.8 keV (C) 0.63 keV (D) 0.16 keV

Tardigrade Admin · Accepted Answer

de-Broglie wavelength, λ=( h /√2 m E) ∴ (λ1/λ2) =√(E2/E1) λ1 = Wave length of electron =0 ⋅ 4 × 10-10 m E1 =1 ⋅ 0 keV λ2 =1 ⋅ 0 × 10-10 m E2 =? (0.4 × 10-10/1 × 10-10) =√(E2/1 ⋅ 0 keV ) (4/10) =√(E2/1) (16/100) =(E2/1) E2 =0 ⋅ 16 keV

Q. The de Broglie wavelength of an electron is 0.4×10−10 m when its kinetic energy is 1.0keV. Its wavelength will be 1.0×10−10m, when its kinetic energy is

0.2 keV

0.8 keV

0.63 keV

0.16 keV

de-Broglie wavelength, λ=2mE​h​ ∴λ2​λ1​​=E1​E2​​​ λ1​= Wave length of electron =0⋅4×10−10m E1​=1⋅0keV λ2​=1⋅0×10−10m E2​=? 1×10−100.4×10−10​=1⋅0keVE2​​​ 104​=1E2​​​ 10016​=1E2​​ E2​=0⋅16keV

Q. The de Broglie wavelength of an electron is $0.4 \times 1 0^{- 10}$ m when its kinetic energy is $1.0 k e V$ . Its wavelength will be $1.0 \times 1 0^{- 10} m$ , when its kinetic energy is