The New England Patriots are set to sign two-time Pro Bowler Austin Hooper, a 29-year-old tight end who last played with the Las Vegas Raiders. The veteran is expected to ink a one-year deal with New England worth up to $4.25 million, per Ian Rapoport of NFL.com.
Hooper again will join an Alex Van Pelt-led offense, as the two worked hand-in-hand during the 2020 and 2021 seasons on the Cleveland Browns. The 6-foot-4 tight end logged 780 yards on 84 receptions with seven touchdowns across both seasons in Cleveland.
The 2016 third-round pick will best be remembered for his time with the Atlanta Falcons, where he achieved back-to-back Pro Bowl seasons in 2018 and 2019, logging 2,244 yards on 214 receptions with 16 touchdowns in his four years with the team.
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In his nine seasons in the league, Hooper has failed to find a true home, bouncing around five different organizations. After being targeted just 31 times last season with the Raiders, converting them into 25 catches for 234 yards, Hooper can be hopeful that a depleted Patriots offense -- now without DeVante Parker -- can give the career journeyman more looks.
While Hooper's numbers have declined since he left Atlanta -- 2023 was his first season without a touchdown -- the Stanford product will now have an opportunity to resurrect his career as the second option behind recently re-signed Hunter Henry.