Question

The work function for tungsten and sodium are 4.5 eV and 2.3 eV respectively. If the threshold wavelength $\lambda$ for sodium is $5460\stackrel{\circ }{A}$ the value of $\lambda$ for tungsten is

• A. $5893\stackrel{\circ }{A}$
• B. $10683\stackrel{\circ }{A}$
• C. $2791\stackrel{\circ }{A}$
• D. $528\stackrel{\circ }{A}$

Answer 1

The correct option is C

$2791\stackrel{\circ }{A}$

Since $W_0=\frac{hc}{{\lambda }_0};\therefore \frac{(W_0{)}_T}{(W_0{)}_{Na}}=\frac{{\lambda }_{Na}}{{\lambda }_T}or{\lambda }_T=\frac{{\lambda }_{Na}\times (W_0{)}_{Na}}{(W_0{)}_T}=\frac{5460\times 2.3}{4.5}=2791\stackrel{\circ }{A}$