Question

The photoelectric threshold wavelength for potassium having work function of 2 eV, is
( Take h = $ 6.6 \times 10^{ - 34} \, J-s, \, 1 \, e V = 1.6 \times 10^{ - 19 } \, J, \, c = 3 \times 10^8 \, ms^{ - 1} )$

• A. 1860 nm
• B. 618.7 nm
• C. 1240 nm
• D. 317.0 nm

Answer 1

Answer: (B)

We know that, work function
$ \phi = h v_0 $
$\Rightarrow \phi = \frac{ hc }{ \lambda_0 } $ $ \bigg [ \because v_0 = \frac{ c }{ \lambda_0 } \bigg ] $
$\Rightarrow \lambda_0 = \frac{ h c }{ \phi } $
$ \therefore \, = \frac{ 6.6 \times 10^{ - 34} \times 3 \times 10^8 }{ 2 \times 1.6 \times 10^{ - 19 }} $
= $ \frac{ 6.6 \times 3 \times 10^{ - 7 }}{ 2 \times 1.6 } $
= $ 6.187 \times 10^{ - 7 } $
= 618.7 nm