The work function of a metallic surface is 5.01eV. The photo-electrons are emitted when light of wavelength 2000 mathringA falls on it. The potential difference applied to stop the fastest photo-electrons is:- [h = 4 . 14 × 10- 15 e V sec]

Question

The work function of a metallic surface is 5.01eV. The photo-electrons are emitted when light of wavelength 2000 mathringA falls on it. The potential difference applied to stop the fastest photo-electrons is:- [h = 4 . 14 × 10- 15 e V sec] (A) 1.2 volts (B) 2.24 volts (C) 3.6 volts (D) 4.8 volts

Tardigrade Admin · Accepted Answer

Energy of incident light E=(12375/2000)=6.18eV According to relation E=W0+eV0 ⇒ V0=((E - W0)/e)=((. 6 . 18 e V - 5 . 01 e V .)/e)=1.17V≈1.2V

Q. The work function of a metallic surface is $5.01 e V .$ The photo-electrons are emitted when light of wavelength $2000 \overset{˚}{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is:-
$[h = 4.14 \times 1 0^{- 15} e V sec]$

$1.2$ volts

$2.24$ volts

$3.6$ volts

$4.8$ volts

Energy of incident light $E = \frac{12375}{2000} = 6.18 e V$
According to relation $E = W_{0} + e V_{0}$
$\Rightarrow V_{0} = \frac{( E - W _{0} )}{e} = \frac{( 6.18 e V - 5.01 e V )}{e} = 1.17 V \approx 1.2 V$

Q. The work function of a metallic surface is 5.01eV. The photo-electrons are emitted when light of wavelength 2000A˚ falls on it. The potential difference applied to stop the fastest photo-electrons is:- [h=4.14×10−15eVsec]

1.2 volts

2.24 volts

3.6 volts

4.8 volts