Question

The de Broglie's wavelength of electron emitted by metal will be ______ Å whose threshold frequency is 2.25 × 10¹⁴ Hz when exposed to visible radiation of wavelength 500 nm.

पर्याय

• 9.65

• 9.84

• 2.56

• 9.03

Answer 1

The de Broglie's wavelength of electron emitted by metal will be 9.84 Å whose threshold frequency is 2.25 × 10¹⁴ Hz when exposed to visible radiation of wavelength 500 nm.

Explanation:

K.E. of ejected electron = hυ – hυ°

but `lambda = h/(mv) = h/(sqrt(2mE))`

E = `[(hc)/lambda - h (2.25 xx 10^14)]`

`lambda = h/(sqrt((h)(2m)(c/lambda - ν_0)))`

`= sqrt(h/((2m)(c/lambda - ν_0)))`

`= sqrt((6.625 xx 10^-34)/((2 xx 9.109 xx 10^-31)(3.75 xx 10^14)))`

= 9.84 × 10^-10 m

= 9.84 Å