When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \mathrm{~V}$. This potential drops to $0.6 \mathrm{~V}$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $\frac{h c}{e}=1.24 \times$ $10^{-6} \mathrm{JmC}^{-1}$.]

_{0}) are given below:

$$\lambda \left( {\mu m} \right)$$ | V_{0}(Volt) |
---|---|

0.3 | 2.0 |

0.4 | 1.0 |

0.5 | 0.4 |

Given that c = 3 $$ \times $$ 10

^{8}ms

^{-1}and e = 1.6 $$ \times $$ 10

^{-19}C, Planck's constant (in units of J-s) found from such an experiment is) :

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u_{1} and u_{2}, respectively. If the ratio u_{1} : u_{2} = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly