Question

If the threshold wavelength of light for photoelectric emission to take place from a metal surface is \( 6250 \, \text{\AA} \), then the work function of the metal is (Planck's constant \( = 6.6 \times 10^{-34} \, \text{Js} \))

• A. 3.98 eV
• B. 1.98 eV
• C. 2.98 eV
• D. 4.98 eV

Hint:

- Work function \( \phi_0 = hf_0 = \frac{hc}{\lambda_0} \), where \(f_0\) is threshold frequency, \( \lambda_0 \) is threshold wavelength. - \(h\) is Planck's constant, \(c\) is speed of light. - Energy in Joules can be converted to eV: \( E(\text{eV}) = \frac{E(\text{J})}{e} \), where \( e \approx 1.602 \times 10^{-19} \, \text{C} \). - Useful approximation: \( E(\text{eV}) \approx \frac{12400}{\lambda(\text{in Angstroms})} \) or \( E(\text{eV}) \approx \frac{1240}{\lambda(\text{in nm})} \). The precise value depends on the \(h,c,e\) values used. For \(h=6.6 \times 10^{-34}\), \(c=3 \times 10^8\), \(e=1.6 \times 10^{-19}\), the factor is \(12375\).

Answer 1

The Correct Option is B

Threshold wavelength \( \lambda_0 = 6250 \, \text{\AA} = 6250 \times 10^{-10} \, \text{m} \).
Planck's constant \( h = 6.
6 \times 10^{-34} \, \text{Js} \).
Speed of light \( c = 3 \times 10^8 \, \text{m/s} \).
The work function \( \phi_0 \) is related to the threshold wavelength \( \lambda_0 \) by: \[ \phi_0 = \frac{hc}{\lambda_0} \] Substitute the values: \[ \phi_0 = \frac{(6.
6 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{6250 \times 10^{-10} \, \text{m}} \] \[ \phi_0 = \frac{6.
6 \times 3}{6250} \times \frac{10^{-34} \times 10^8}{10^{-10}} \, \text{J} \] \[ \phi_0 = \frac{19.
8}{6250} \times 10^{-26+10} \, \text{J} = \frac{19.
8}{6250} \times 10^{-16} \, \text{J} \] \[ \phi_0 = \frac{19.
8}{6.
25 \times 10^3} \times 10^{-16} = \frac{19.
8}{6.
25} \times 10^{-19} \, \text{J} \] \( \frac{19.
8}{6.
25} = \frac{1980}{625} \).
\( 1980 \div 625 \approx 3.
168 \).
So, \( \phi_0 = 3.
168 \times 10^{-19} \, \text{J} \).
To convert Joules to electronvolts (eV), divide by the charge of an electron \( e \approx 1.
6 \times 10^{-19} \, \text{C} \).
\( (1 \text{ eV} = 1.
6 \times 10^{-19} \text{ J}) \).
\[ \phi_0 (\text{in eV}) = \frac{3.
168 \times 10^{-19} \, \text{J}}{1.
6 \times 10^{-19} \, \text{J/eV}} = \frac{3.
168}{1.
6} \, \text{eV} \] \[ \frac{3.
168}{1.
6} = \frac{31.
68}{16} \).
\( 31.
68 \div 16 \): \( 16 \times 1 = 16 \), remainder \( 15.
6 \).
\( 16 \times 0.
9 = 14.
4 \), remainder \( 1.
28 \).
\( 16 \times 0.
08 = 1.
28 \).
So, \( \frac{31.
68}{16} = 1.
98 \).
\[ \phi_0 = 1.
98 \, \text{eV} \] This matches option (2).
A useful shortcut: \( E(\text{eV}) = \frac{12400}{\lambda(\text{\AA})} \) or \( E(\text{eV}) = \frac{12375}{\lambda(\text{\AA})} \) (using more precise \(h,c,e\)).
Using \( \frac{12400}{6250} = \frac{1240}{625} \).
\( 1240 \div 625 \): \( 625 \times 1 = 625 \).
\( 1240 - 625 = 615 \).
Decimal point.
\( 6150 \div 625 \).
\( 625 \times 9 = 5625 \).
\( 6150 - 5625 = 525 \).
Decimal.
\( 5250 \div 625 \).
\( 625 \times 8 = 5000 \).
So, \( \approx 1.
98.
.
.
\) eV.
The calculation used \(hc = 19.
8 \times 10^{-26}\).
Using \(12400/\lambda(\text{A})\) uses \(h=6.
626 \times 10^{-34}, c=2.
998 \times 10^8, e=1.
602 \times 10^{-19}\).
The values given \(h=6.
6 \times 10^{-34}\) and implicitly \(c=3 \times 10^8\) and \(e=1.
6 \times 10^{-19}\) lead to \( \frac{12375}{\lambda(\text{A})} \).
If \(hc/e\) is approx 1240 nm.
eV, then \(12400\) A.
eV.
\( \frac{6.
6 \times 3 \times 10^{-26}}{1.
6 \times 10^{-19}} = \frac{19.
8}{1.
6} \times 10^{-7} = 12.
375 \times 10^{-7} \) Jm/C.
For eV.
A use: \( E(eV) = \frac{hc(Jm)}{\lambda(m) e(C)} \).
\( hc = 19.
8 \times 10^{-26} \text{ Jm} \).
\( \lambda_0 = 6250 \times 10^{-10} \text{ m} \).
\( \phi_0 = \frac{19.
8 \times 10^{-26}}{6250 \times 10^{-10} \times 1.
6 \times 10^{-19}} = \frac{19.
8}{6250 \times 1.
6} \frac{10^{-26}}{10^{-29}} = \frac{19.
8}{10000} \times 10^3 = \frac{19.
8}{10} = 1.
98 \) eV.